1.Contest2425 - 毕老师算法试验一
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题目总览:
A.排列问题
// 排列问题// 输出一个可能含有反复字符的字符串，打印出该字符串中所有字符的全排列。#include<bits/stdc++.h>using namespace std;bool IsSwap(vector<char> &chars, int nBegin, int nEnd){ for(int i=nBegin; i<nEnd; i++) //记住是<nEnd { if(chars[i] == chars[nEnd]) return false; } return true;}void Perm(vector<char> &chars, int k, int m){ if(k==m) { for(const char x:chars) { cout<<x; } cout<<endl; } else { for(int i=k; i<=m; i++) { if(IsSwap(chars, k, i)) { swap(chars[k], chars[i]); Perm(chars, k+1, m); swap(chars[k], chars[i]); } } }}void Foo(vector<char> &chars){ Perm(chars, 0, chars.size()-1);}// 字符串bool IsSwap(string str, int nBegin, int nEnd){ for(int i=nBegin; i<nEnd; i++) //记住是<nEnd { if(str[i] == str[nEnd]) return false; } return true;}void Perm2(string str, int k, int m){ if(k==m) { cout<<str; cout<<endl; } else { for(int i=k; i<=m; i++) { if(IsSwap(str, k, i)) { swap(str[k], str[i]); Perm2(str, k+1, m); swap(str[k], str[i]); } } }}void Foo2(string str){ str.pop_back(); cout<<str<<endl; Perm2(str, 0, str.length()-1);}int main(){ // vector<char> chars; // int n; // cin>>n; // char ele; // for(int i=0; i<n; i++) // { // cin>>ele; // chars.push_back(ele); // } // // for(const char x:chars) cout<<x; // Foo(chars); string str; // scanf("%s", str); https://blog.csdn.net/liu16659/article/details/86772657 不倡议应用scanf输出string类型 cin>>str; Foo2(str); system("pause"); return 0;}B.疾速幂
// B.疾速幂#include<bits/stdc++.h>using namespace std;const long long Mod = 1e8 + 7;typedef long long ll;ll Pow(ll base, ll power){ ll result = 1; for(int i=1; i<=power; ++i) { result *= power; result %= 1000; //每步后果提前进行取模 } return result%1000;}// 疾速幂// 所以咱们疾速幂算法的核心思想就是每一步都把指数分成两半，// 而相应的底数做平方运算。这样不仅能把十分大的指数给一直变小，// 所须要执行的循环次数也变小，而最初示意的后果却始终不会变ll fastPow(ll base, ll power){ ll result = 1; while(power>0) { if(power%2 == 0) { base = base*base % Mod; power /= 2; } else { power -= 1; result = result*base % Mod; base = base*base % Mod; power /= 2; } } return result;}ll myPow(int x){ ll result = 0; for(int i=1; i<=x; ++i) result += fastPow(i, i); return (result+1)%Mod;}int main(){ int x; while(cin>>x) { cout<<myPow(x)<<endl; } system("pause"); return 0;}C.求第k小的数
// C.求第k小的数#include<bits/stdc++.h>using namespace std;int Partition(int a[], int low, int high);template<class Type>int RandomizedPartition(Type a[], int p, int r);template<class Type>void RandomizedQuickSort(Type a[], int p, int r);template<class Type>Type RandomizedSelect(Type a[], int p, int r, int k);int main(){ // int n, k; // cin>>n>>k; // vector<int> vec; // for(int i=0; i<n; ++i) // { // int num; // scanf("%d", &num); // vec.push_back(num); // } // sort(vec.begin(), vec.end()); // cout<<vec[k-1]; int a[100], n; while(~scanf("%d", &n)) { memset(a, 0, sizeof(a)); for(int i=0; i<n; i++) scanf("%d",&a[i]); int k; scanf("%d", &k); cout<<RandomizedSelect(a, 0, n-1, k)<<endl; } system("pause"); return 0;}// 2.RandomizedSelect,均匀工夫复杂度O(n)int Partition(int a[], int low, int high){ int key = a[low]; while(low<high) { while(low<high&&a[high]>key) --high; a[low] = a[high]; while(low<high&&a[low]<=key) ++low; a[high] = a[low]; } a[low] = key; return low;}template<class Type>int RandomizedPartition(Type a[], int p, int r){ int i = rand()%(r-p+1) + p; swap(a[i], a[p]); return Partition(a, p, r);}// template<class Type>// void RandomizedQuicksort(Type a[], int p, int r) //随机快排// {// if(p<r)// {// int q = RandomizedPartition(a, p, r);// RandomizedQuicksort(a, p, q-1); //对左半段排序// RandomizedQuicksort(a, q+1, r); //对右半段排序// }// }template<class Type>Type RandomizedSelect(Type a[], int p, int r, int k) //随机划分,抉择{ if(p == r) return a[p]; int loc = RandomizedPartition(a, p ,r); int count = loc-p+1; //count为a[p:r]中的元素个数 if(k<=count) return RandomizedSelect(a, p, loc, k); else return RandomizedSelect(a, loc+1, r, k-count);}D.外部收益率
// D.外部收益率#include<bits/stdc++.h>using namespace std;int main(){ int a[100], n; while(scanf("%d", &n) && n) { for(int i=0; i<=n; i++) scanf("%d", a+i); double x = -1.0, y=1e6, irr, npv; for(int j=0; j<200; j++) { irr = (x+y)/2; //枚举irr的取值范畴[-1.0, 1e6], 二分法迫近 (至于这个范畴怎么来的,我就不分明了.数学问题) npv = 0; for(int k=0; k<=n; k++) npv += 1.0*a[k]/pow(1+irr, k); if(fabs(npv)<1e-6) break; //小于一个阈值,认为就是方程的解 if(npv < 0) y = irr; if(npv > 0) x = irr; } printf("%.2lf\n", irr); memset(a, 0, sizeof(a)); //每次重置a数组 } // cout<<9/2.25<<endl; system("pause"); return 0;}E.跳台阶
// 跳台阶#include<bits/stdc++.h>using namespace std;// f(x) = f(x-1) + f(x-2)int climbStairsMemo(int n, vector<int>& memo){ if(memo[n]>0) return memo[n]; //记录数组 if(n==1 || n==2) memo[n] = n; else memo[n] = climbStairsMemo(n-1, memo) + climbStairsMemo(n-2, memo); return memo[n];}int climbStairs(int n) { vector<int> memo(n+1, 0); return climbStairsMemo(n, memo);}// 滚动数组 将空间复杂度由O(n)优化成O(1)int climbStairs2(int n){ int p=0, q=0, r=1; for(int i=1; i<=n; i++) { p = q; q = r; r = p+q; } return r;}int main(){ int n; cin>>n; cout<<climbStairs2(n)%1000000007<<endl; system("pause"); return 0;}2.Contest2445 - 毕老师算法试验二
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题目总览:
A.沙子的品质
// A.沙子的品质 OJ作业一D题// 动静布局#include<bits/stdc++.h>using namespace std;const int inf = 0x3f3f3f3f; //为什么用0x3f3f3f3f而不必0xffffff https://blog.csdn.net/jiange_zh/article/details/50198097int Compute(int *nums, int n){ int Sum[n+1]={0}; int DP[n+1][n+1]; //之前用DP[1005][1005],导致栈空间溢出 https://blog.csdn.net/kangyupl/article/details/90723367 memset(DP, inf, sizeof(DP)); for(int i=1; i<=n; i++) { Sum[i] = Sum[i-1] + nums[i]; DP[i][i] = 0; } for(int Len=2; Len<=n; Len++) for(int left=1; left<=n-Len+1; left++) { int right = left+Len-1; for(int k=left; k<right; k++) { DP[left][right] = min(DP[left][k]+DP[k+1][right]+Sum[right]-Sum[left-1], DP[left][right]); } } return DP[1][n];}int main(){ int n; cin>>n; int nums[n+1]; nums[0] = 0; for(int i=1; i<=n; ++i) cin>>nums[i]; cout<<Compute(nums, n)<<endl; system("pause"); return 0;}B.最长公共子序列
// B.最长公共子序列 OJ作业一F题// 动静布局#include<bits/stdc++.h>using namespace std;int longestCommonSequence(string a, string b){ int M = a.size(); int N = b.size(); vector<vector<int>> DP(M+1, vector<int>(N+1, 0)); for(int i=1; i<=M; ++i) { for(int j=1; j<=N; ++j) { if(a[i-1]==b[j-1]) DP[i][j] = DP[i-1][j-1] + 1; else DP[i][j] = max(DP[i-1][j], DP[i][j-1]); } } return DP[M][N];}int main(){ string a, b; cin>>a>>b; cout<<longestCommonSequence(a, b)<<endl; system("pause"); return 0;}C.三角形的门路权
// C.三角形最小门路权// Leetcode 125题// 动静布局#include<bits/stdc++.h>using namespace std;// 了解谬误,直角三角形的状况.int minimumTotal2(vector<vector<int>> &triangle){ int n = triangle.size(); vector<vector<int>> dp(n, vector<int>(n)); dp[0][0] = triangle[0][0]; dp[1][0] = triangle[1][0], dp[1][1] = triangle[0][0]+triangle[1][1]; if(n==2) return dp[1][1]; for(int i=2; i<n; i++) { dp[i][0] = dp[i-1][1] + triangle[i][0]; for(int j=1; j<=i-2; j++) { dp[i][j] = max(dp[i-1][j-1], dp[i-1][j+1]) + triangle[i][j]; } dp[i][i] = dp[i-1][i-1] + triangle[i][i]; dp[i][i-1] = dp[i-1][i-2] + triangle[i][i-1]; } cout<<endl<<endl<<"DP数组更新为:"<<endl; for(const auto x:dp) { for(const int y:x) cout<<y<<" "; cout<<endl; } return *max_element(dp[n-1].begin(), dp[n-1].end());}// 看成等腰三角形int maximumTotal(vector<vector<int>> &triangle){ int n = triangle.size(); vector<vector<int>> dp(n, vector<int>(n,0)); dp[0][0] = triangle[0][0]; for(int i=1; i<n; i++) { dp[i][0] = dp[i-1][0] + triangle[i][0]; for(int j=1; j<n-1; j++) { dp[i][j] = max(dp[i-1][j-1], dp[i-1][j]) + triangle[i][j]; } dp[i][i] = dp[i-1][i-1] + triangle[i][i]; } return *max_element(dp[n-1].begin(), dp[n-1].end());}int main(){ int n; cin>>n; vector<vector<int>> triangle(n, vector<int>(n)); for(int i=0; i<n; i++) { for(int j=0; j<=i; j++) cin>>triangle[i][j]; } cout<<maximumTotal(triangle)<<endl; system("pause"); return 0;}D.跳跃游戏Ⅱ
// D.跳跃游戏Ⅱ // LeetCode第45题// 贪婪算法#include<bits/stdc++.h>using namespace std;// 思维就一句话：每次在上次能跳到的范畴（end）内抉择一个能跳的最远的地位（也就是能跳到max_far地位的点）作为下次的起跳点 ！int Jump(vector<int> &nums){ int n = nums.size(); int max_far = 0; //目前可能跳到的最远地位 int step = 0; int end = 0; //上次跳跃可达到的范畴右边界 for(int i=0; i<n-1; i++) //<n-1,最初一个元素不必遍历. { max_far = max(i+nums[i], max_far); if(i == end) { end = max_far; step++; } } return step;}int main(){ int n; cin>>n; vector<int> nums(n); for(int i=0; i<n; i++) cin>>nums[i]; cout<<Jump(nums)<<endl; system("pause"); return 0;}E.字母排序
// E.字母排序// 最长不降子序列#include<bits/stdc++.h>using namespace std;// 1.LIS ---> LCS 最长递增子序列(排序+LCS)// const int N = 1010;// int DP[N][N];// int DIR[N][N];// int LCS_Length(string a, string b)// {// int m = a.size();// int n = b.size();// for(int i=1; i<=m; i++)// {// for(int j=1; j<=n; j++)// {// if(a[i-1] == b[j-1])// {// DP[i][j] = DP[i-1][j-1] + 1;// DIR[i][j] = 1;// }// else if(DP[i-1][j] > DP[i][j-1])// {// DP[i][j] = DP[i-1][j];// DIR[i][j] = 2;// }// else // {// DP[i][j] = DP[i][j-1];// DIR[i][j] = 3;// }// }// }// return DP[m][n];// }// void LCS(string a, int i, int j)// {// if(i==0 || j==0) return;// if(DIR[i][j] == 1)// {// LCS(a, i-1, j-1);// cout<<a[i-1];// }// else if(DIR[i][j] == 2) LCS(a, i-1, j);// else if(DIR[i][j] == 3) LCS(a, i, j-1);// }// string RD(string b, int &add) //去重函数 hhh变为h// {// bool Flag[b.size()] = {false};// for(int i=1; i<b.size(); i++)// {// if(b[i] == b[i-1]) // {// add++;// Flag[i] = true;// }// }// for(int i=0; i<b.size(); i++)// {// if(Flag[i] == 1) b[i] = '.';// }// // cout<<b<<endl;// b.erase(std::remove(b.begin(), b.end(), '.'), b.end()); //去重操作// return b;// }// int LIS_Length(string a, string b)// {// return LCS_Length(a, b);// }// int main()// {// string a, str;// cin>>a;// cin.ignore(); //空格// getline(cin, str);// str += ' ';// string b;// int Start_pos = 0;// int add = 0;// for(int i=0; i<str.size(); i++)// {// if(str[i] == ' ')// {// // cout<<"Flag"<<i<<endl;// b = str.substr(Start_pos, i-Start_pos);// // cout<<b<<endl;// RD(b, add);// cout<<LIS_Length(a, b)+add;// Start_pos = i+1;// add = 0;// }// }// // string test;// // cin>>test;// // int add = 0;// // cout<<RD(test, add)<<" "<<add<<endl;// // cout<<test<<endl;// system("pause");// return 0;// }// 2.DP// O(n^2)// int LengthOfLIS(vector<int> &nums)// {// int n = nums.size();// if(n==0) return 0;// vector<int> dp(n, 1);// for(int i=0; i<n; i++)// {// for(int j=0; j<i; j++)// {// if(nums[i] > nums[j])// {// dp[i] = max(dp[j]+1, dp[i]);// }// }// }// return *max_element(dp.begin(), dp.end());// }// int main()// {// int n;// while(scanf("%d", &n) && n)// {// vector<int> nums(n);// for(int i=0; i<n; i++) scanf("%d", &nums[i]);// cout<<LengthOfLIS(nums)<<endl;// }// system("pause");// return 0;// }// 3.贪婪 + 二分查找// 保护一个一维数组B，并且这个数组是动静扩大的，初始大小为1，//B[i]示意最长回升子序列长度是i的所有子串中开端最小的那个数，依据这个数字，咱们能够比拟晓得，//只有以后考查的这个数比B[i]大，那么以后这个数肯定能通过B[i]形成一个长度为i+1的回升子序列。//如果以后考查的数并没有数组尾部的B[i]大,那么咱们抉择在B数组中找到第一个大于以后考查数的元素,并且更换.// 正确!!!int BinarySearch(vector<int> &nums, int l, int r, int target) //在nums数组中找到第一个大于target的数,而后更新它{ if(l == r) return l; int mid; while(l <= r) { mid = (l+r)/2; if(nums[mid] == target) return mid; //return mid! else if(nums[mid] > target) r = mid-1; else l = mid+1; } return l;}int LengthOfLIS(vector<int> &nums){ int n = nums.size(); vector<int> B(n+1); B[0] = 0; B[1] = nums[0]; int len = 1; //示意B数组的长度 for(int i=1; i<n; i++) { if(nums[i] > B[len]) { B[++len] = nums[i]; } else if(nums[i] < B[len]) { int pos = BinarySearch(B, 1, len, nums[i]); //在B数组中找到第一个大于nums[i]的元素地位 cout<<pos<<" "<<i<<" "<<nums[i]<<endl; B[pos] = nums[i]; } } for(const int x:B) cout<<x<<" "; cout<<endl; return len;}// 力扣! https://leetcode-cn.com/problems/longest-increasing-subsequence/solution/zui-chang-shang-sheng-zi-xu-lie-by-leetcode-soluti/// 思考一个简略的贪婪，如果咱们要使回升子序列尽可能的长，则咱们须要让序列回升得尽可能慢，因而咱们心愿每次在回升子序列最初加上的那个数尽可能的小。// 咱们顺次遍历数组 nums 中的每个元素，并更新数组 d 和 len 的值。如果nums[j] > d[len] 则更新 len = len + 1// 否则在 d[1...len] 中 找到 d[i-1] < nums[j] < d[i]的pos值,而后更新d[i] = nums[j]int lengthOfLIS(vector<int> &nums){ int len = 1, n = int(nums.size()); if(n == 0) return 0; vector<int> d(n+1, 0); d[len] = nums[0]; for(int i=1; i<n; ++i) { if(nums[i] > d[len]) d[++len] = nums[i]; else { int l = 1, r = len, pos = 0; //如果找不到阐明所有的数都比nums[i]大,此时要更新d[1] while(l <= r) { int mid = (l+r) >> 1; if(d[mid] < nums[i]) { pos = mid; l = mid+1; } else r = mid-1; } d[pos+1] = nums[i]; } } return len;}int main(){ int t; cin>>t; while(t--) { int n; cin>>n; vector<int> nums(n); for(int i=0; i<n; i++) cin>>nums[i]; cout<<LengthOfLIS(nums)<<endl; } system("pause"); return 0;}3.Contest2453 - 毕老师算法试验三
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题目总览:
A.Homework
// D.Homework// 正确!#include<bits/stdc++.h>using namespace std;bool cmp(pair<double, double> a, pair<double, double> b){ return a.second/a.first > b.second/b.first;}int main(){ int M, N; while(scanf("%d%d", &M, &N) && (M && N)) { vector<pair<double, double>> vec; double time, value; for(int i=0; i<M; ++i) { scanf("%lf%lf", &time, &value); vec.push_back(pair<double, double>(time, value)); } sort(vec.begin(), vec.end(), cmp); double TimeTotal = N; double ValueTotal = 0; //double类型,如果写为int类型答案谬误 for(const auto x:vec) cout<<"("<<x.first<<","<<x.second<<")"<<endl; for(const pair<double, double> x:vec) { if(TimeTotal-x.first >= 0) { cout<<"Flag1"<<endl; ValueTotal += x.second; TimeTotal -= x.first; } else { ValueTotal += TimeTotal * x.second/x.first; break; } } printf("%.2lf\n", ValueTotal); } system("pause"); return 0;}B.区间蕴含问题
// E.区间蕴含问题// 贪婪算法 相似于流动安顿问题#include<bits/stdc++.h>using namespace std;const int maxn = 10010;struct Node{ int left, right;}node[maxn];bool cmp(Node a, Node b){ return a.right < b.right;}int main(){ int n, m; scanf("%d %d", &n, &m); for(int i=0; i<n; ++i) { scanf("%d %d", &node[i].left, &node[i].right); // node[i].right++; } sort(node, node+n, cmp); int l, r; for(int i=0; i<m; ++i) { scanf("%d %d", &l, &r); int pos = l, ans = 0; for(int j=0; j<n; ++j) { if(node[j].left>=pos && node[j].right<=r) { cout<<l<<" "<<r<<endl; cout<<j<<" "<<node[j].left<<" "<<node[j].right<<endl; ans++; pos = node[j].right; } } printf("%d\n", ans); } system("pause"); return 0;}C.最长子序列
// F.最长子序列(最大子段和)#include<bits/stdc++.h>using namespace std;// 1.递归分治法// f(i) = max(f(i-1)+nums[i], nums[i]) 思考f(i-1)带来的是正能量还是负能量// f(i) 代表以第 i 个数结尾的「间断子数组的最大和」，那么很显然咱们要求的答案就是：// max(0-i-1){f[i]}int maxSubArray(vector<int> &nums){ int n = nums.size(); if(n==0) return {}; if(n==1) return nums[0]; vector<int> f(n); f[0] = nums[0]; for(int i=1; i<n; ++i) { f[i] = max(f[i-1]+nums[i], nums[i]); } for(const int x:f) cout<<x<<" "; cout<<endl; return *max_element(f.begin(), f.end()); //记住不是返回f[n-1]}int maxSubArray2(vector<int> &nums){ int n = nums.size(); if(n==0) return {}; int pre = 0, maxAns = nums[0]; for(const int x:nums) { pre = max(pre+x, x); //pre记录以后x后面的间断子数组的最大和,即f[i-1],不断更新. 相似于滚动数组 maxAns = max(maxAns, pre); } return maxAns;}int main(){ int n; while(~scanf("%d", &n)) { vector<int> nums(n); for(int i=0; i<n; ++i) scanf("%d", &nums[i]); cout<<maxSubArray(nums)<<endl; } system("pause"); return 0;}D.三值排序
// D.三值排序#include<bits/stdc++.h>using namespace std;int ThreeSort(vector<int> &nums){ int n = nums.size(); int a=0, b=0, c=0; for(int i=0; i<n; ++i) { if(nums[i]==1) a++; else if(nums[i]==2) b++; else if(nums[i]==3) c++; } int dif[4][4] = {0}; for(int i=0; i<a; i++) if(nums[i] != 1) dif[1][nums[i]]++; for(int i=a; i<a+b; i++) if(nums[i] != 2) dif[2][nums[i]]++; for(int i=a+b; i<n; i++) if(nums[i] != 3) dif[3][nums[i]]++; int ans = 0, sum = 0; for(int i=1; i<4; i++) { for(int j=i+1; j<4; j++) { if(dif[i][j] <= dif[j][i]) { ans += dif[i][j]; sum += dif[j][i] - dif[i][j]; } else { ans += dif[j][i]; sum += dif[i][j] - dif[j][i]; } } } ans = ans + sum/3*2; //三个数字都不在本人该在的地位，然而只须要两步就能变得有序：213变为123只需两步 312 132等等同样 return ans;}int main(){ int n; cin>>n; vector<int> nums(n); int a=0, b=0, c=0; for(int i=0; i<n; ++i) scanf("%d", &nums[i]); cout<<ThreeSort(nums)<<endl; system("pause"); return 0;}E.法师康的工人
// E.法师康的工人// 这一题有点问题,我如同没改过来#include<bits/stdc++.h>using namespace std;bool cmp(pair<int, int> a, pair<int, int> b){ return a.second < b.second;}int main(){ int n; cin>>n; vector<pair<int, int>> vec; int left, right; for(int i=0; i<n; ++i) { scanf("%d%d", &left, &right); vec.push_back(pair<int, int>(left, right)); } sort(vec.begin(), vec.end(), cmp); int left_border = vec[0].first; int right_border = vec[0].second; int max_ContinuousLength = right_border-left_border; int max_IntervalLength = 0; for(int i=1; i<n; ++i) { if(vec[i].first <= right_border) { right_border = max(right_border, vec[i].second); left_border = min(left_border, vec[i].first); } else { max_ContinuousLength = max(max_ContinuousLength, right_border-left_border); max_IntervalLength = max(max_IntervalLength, vec[i].first-right_border); left_border = vec[i].first; //呈现距离,重置边界 right_border = vec[i].second; } } max_ContinuousLength = max(max_ContinuousLength, right_border-left_border); // 避免 2 (100 500) (400 900)这种没有间断的状况呈现,间接跳出循环的话 cout<<max_ContinuousLength<<" "<<max_IntervalLength<<endl; system("pause"); return 0;}// 10// 2 3// 4 5// 6 7// 8 9// 10 11// 12 13// 14 15// 16 17// 18 19// 1 20// 19 1// 请按任意键持续. . .// 正确答案:19 0 谬误的起因?咋改4.Contest2462 - 毕老师算法试验四
传送门在这里
题目总览:
A.凯撒加密法
// A.żČöĂÜÂë#include<bits/stdc++.h>using namespace std;string CaesarCipher(string ciphertext, int n){ int length = ciphertext.size(); for(int i=0; i<length; ++i) { if(isupper(ciphertext[i])) ciphertext[i] = 'A' + (ciphertext[i]-'A'-n+26)%26; else if(islower(ciphertext[i])) ciphertext[i] = 'a' + (ciphertext[i]-'a'-n+26)%26; } return ciphertext;}int main(){ int T; scanf("%d", &T); // for(int i=0; i<T; ++i) // { // string ciphertext; // // cin>>ciphertext; // cin.ignore(); // getline(cin, ciphertext); // int k; // cin>>k; // k = k%26; // cout<<CaesarCipher(ciphertext, k)<<endl; // } string str[T]; int change[T]; for(int i=0; i<T; ++i) { cin>>str[i]; cin>>change[i]; change[i] %= 26; } for(int i=0; i<T; i++) { cout<<CaesarCipher(str[i], change[i])<<endl; } system("pause"); return 0;}B.维吉尼亚明码
// B.¸ĽźŞÄáŃÇĂÜÂë#include<bits/stdc++.h>using namespace std;string fill(string key, string plainText){ int len1 = key.size(); int len2 = plainText.size(); if(len1 < len2) { int num = len2 - len1; string temp = key; for(int i=0; i<num; ++i) { key += temp[i%len1]; } } for(int i=0; i<len2; ++i) { if(islower(key[i])) key[i] -= 32; } return key;}int main(){ string key, cipherText; cin>>key>>cipherText; key = fill(key, cipherText); // cout<<key<<endl; int len = cipherText.size(); string plainText; int change; for(int i=0; i<len; ++i) { change = key[i] - 'A'; // cout<<i<<": "<<change<<endl; if(isupper(cipherText[i])) plainText.push_back('A' + (cipherText[i]-'A'- change + 26)%26); else plainText.push_back('a' + (cipherText[i]-'a'- change + 26)%26); } cout<<plainText<<endl; system("pause"); return 0;}C.简略的明码
#include<bits/stdc++.h>using namespace std;int main(){ int n; int f[35] = {0, 0, 0, 1}; for(int i=4; i<=30; ++i) f[i] = f[i-1]*2 + pow(2, i-4) - f[i-4]; while(cin>>n && n>=1 && n<=30) cout<<f[n]<<endl; system("pause"); return 0;}D.乏味的素数
// 带佬代码#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int n,ans,last;bool visited[32];bool isPrime[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};void pc(int cur){ if(cur == n && isPrime[1+last]) { ans++; return; } for(int i = 2; i <= n; i++) { if(!visited[i] && isPrime[last+i]) { int t = last; last = i; visited[i] = true; pc(cur+1); visited[i] = false; last = t; } }}int main(){ while(~scanf("%d",&n)) { memset(visited,0,sizeof(visited)); ans = 0; last = 1; pc(1); printf("%d\n",ans); } return 0;}E.数据加密
// E.数据加密#include<bits/stdc++.h>using namespace std;string CipherCompute(string PlainText){ int n = PlainText.size(); string ans; if(n==0) return {}; if(n==1) return PlainText; int left = 0, right = n-1; while(n>0) { if(PlainText[left] < PlainText[right]) { ans += PlainText[left]; left++; n--; } else if(PlainText[left] > PlainText[right]) { ans += PlainText[right]; right--; n--; } else { int temp_left = left+1, temp_right = right-1; while(temp_left<n && temp_right>=0 && temp_left <= temp_right && PlainText[temp_left] == PlainText[temp_right]) { temp_left++; temp_right--; } if(PlainText[temp_left]>PlainText[temp_right]) { ans += PlainText[right]; right--; n--; } else { ans += PlainText[left]; left++; n--; } } } return ans;}int main(){ int t; cin>>t; while(t--) { int n; cin>>n; string PlainText; cin>>PlainText; string Cipher = CipherCompute(PlainText); cout<<Cipher<<endl; } system("pause"); return 0;}// 带佬代码#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){ int n; char str[2005],ans[2005]; while(~scanf("%d",&n)) { scanf("%s",str); int i = 0,j = n-1; while(i <= j) { bool flag = false; for(int k = 0; i+k < j-k; k++) { if(str[i+k] < str[j-k]) { flag = true; break; } if(str[i+k] > str[j-k]) break; } if(flag) printf("%c",str[i++]); else printf("%c",str[j--]); } printf("\n"); } return 0;}(～￣▽￣)～
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