Mrrr_Li的博客

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>using namespace std;char c;int t[26],i,j,maxx;int main(){ while(~scanf("%c",&c)) if(c&.

#include<stdio.h>int a[3];char s1,s2;int main(){ while (scanf("%c:=%c;",&s1,&s2)==2) a[s1-'a']=s2>='0' && s2<='9' ? s2-'0' : a[s2-'a']; //赋值语句简洁明了 printf("%d %d %d",a[0],a[1],a[2]); return 0;}...

#include<bits/stdc++.h>using namespace std;char dic[30][20]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty".

#include<bits/stdc++.h>using namespace std;int main(){ string s; char p=0;//放符号 int cnt=0; cin>>s; for(int i=0;i<s.size();i++) { if(s[i]>='0'&&s[i]<='9') cnt++;//记录第一个数长度 else //.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int boy, girl, n; //自动初始化 int main(){ char a[100000]; scanf("%s", a); int n = strlen(a); for(int i = 0; i < n-2.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int isprime(int n){ int k = sqrt(n), i; if ( n <= 1) return 0; else if ( n == 2) return 1; else{ for ( i = 2; i <= k;.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int isprime(int n){ int k = sqrt(n), i; if ( n <= 1) return 0; else if ( n == 2) return 1; else{ for ( i = 2; i <= k;.

C语言学得不精，稍微复杂一点的字符串处理就不会了，在此留下洛谷带佬的C++代码（C++有很多好用的方法），证明我来过（逃#include<iostream>#include<cstring>#include<cstdio>using namespace std;string sent,word;int lens,lenw,t,pos; int main(){ getline(cin,word); getline(cin,sent); .

#include<iostream>#include<cmath>#include<string>#include<fstream>using namespace std;int n,a;string qwq;string c1;string b1;int b,c,d=-1,e;//并不全有用int main(){ cin>>n; cin>>qwq; for(int i=0;i<n;i++) { .

#include <iostream> //cin & cout 用iostream#include <cstring> //memset & strlen 用cstring#include <cstdio> //sscanf & sprintf 用cstdiousing namespace std;int main(){ char a;//a用于存储运算符 int n,c,d;//n存储个数不解释，cd存储两个数字 char s[.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main() { int n; scanf("%d", &n); char arr[110]; scanf("%s", arr); int len = strlen(a.

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=12;int n;struct Matrix{ int a[maxn][maxn]; void in(){for(int i=0;i<n;i++)for(int j=0;j<n;j++)scanf(" %c",a[i]+j);} Matrix rot.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int i, n, num, sum; char text[40000], str[200]; //text：最终字符串，str：缓冲字符串 scanf("%s", str); n = strlen(str);.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>#define f(i,j,n) for(i=j;i<=n;i++) //for循环简写，福利福利~int main(){ int n, i = 0, j, k, sum = 0, arr[40000], cnt = 0; scanf.

这道题数据比较小，打表就完事了，想看正解请移步#include <bits/stdc++.h>using namespace std;int n;int main () { cin >> n; if(n == 1) cout << " 1\n"; if(n == 2) cout <<" 1 2\n 4 3\n"; if(n == 3) { cout << " 1 2 3\n"; cout <&.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>#define f(i,j,n) for(i=j;i<=n;i++) //for循环简写，福利福利~int main(){ int n, t, i, j, ans = 0; double a; scanf("%d", &amp.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int s1, s2, s3, result[100000], i, j, k, temp, max = -1, max2 = -2; scanf("%d %d %d", &s1, &s2, &s3); for.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int m, i, j, k, cnt = 0, sum; scanf("%d", &m); int arr[m], sign[m]; for ( i = 0; i < m; i++) //要做标记，不然可能出现重.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>#define MAX 45int main(){ int n, x, y, num, i, j; scanf("%d", &n); if (n == 1){ printf("1"); return 0; }.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>#define MAX 21int main(){ int i, j, k, w, x, h, q, x1, y1, z1, x2, y2, z2, cnt = 0; scanf("%d %d %d", &w, &x, &h.

写代码首先要逻辑清晰，这道题是先看她能否剩下钱，如果能，再把整100的存给麻麻，零钱给自己。循环每一轮都先判断她能否剩下钱#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int i, arr[12], result = 0; int sum2 = 0; for (

注意这道题是刚好52周，所以直接解方程就行了#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int n, x, k; scanf("%d", &n); for (k = 1;; ++k)//k越小越好，上限没有定 for (x = 100;.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main() { int n, i, j; scanf("%d", &n); for (i = 2; i <= sqrt(n); i++){ if (n%i ==..

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main() { int n, ans = 0, max = -1, i; scanf("%d", &n); int arr[n]; .

写递归函数会超时（⊙﹏⊙b汗，注意ans还要初始化，不然过不去#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>double fei(int n){ int i; double num1 = 1, num2 = 1, ans = 0; if ( n == 1 || n == 2) return 1; e.

写代码5分钟， 改代码两小时因为超时改了快半个小时我也是吐了#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int isprime( int n){ int i, k = sqrt(n); if (n < 2) return 0; else if ( n == 2) return 1; e

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int K, N, coin = 0; scanf("%d", &K); //输入总天数K for(N = 1; K - N >= 0; K -= N++) //第.

Python 3高精度n=eval(input()) #输入一个数字n，记得要用eval(),不然输入的是字符串ans=0 #定义一个数，用来保存答案for i in range(1,n+1): #从1到n来算阶乘 d=1 #阶乘值初始化为1 for j in range(1,i+1): d*=j #简单的阶乘算法 ans+=d #每次都加进去print(ans) #.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int cmp( void const *a, void const *b){ return (*(int*)a) - (*(int*)b);}int gcd(int a, int b){ return b == 0?a:gcd(b, a%b);}int .

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int n, k, ans = 0, i; scanf("%d %d", &n, &k); for( i = 1; i <= k; i++)//要游k天，所以用循环 { i.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ char a[14], mod[12] = "0123456789X"; //先将mod11后的十一个字符存入数组 scanf("%s", a); //洛谷用gets读取不让过 int i, j = 1, t = 0; for.

模拟 高性能AC代码：#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main() { int n, i, j, k = 0, arr[3][2], temp[3] = {0}, min; scanf("%d", &n); for ( i = 0; i < 3; i.

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main() { unsigned long long n, ans; scanf("%lld", &n); ans=n * (n-1) / 2 * (n-2) / 3 * (n-3) / 4; printf("%lld.

代码写得有点乱（逃#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main() { int m, t, s; scanf("%d %d %d", &m, &t, &s); if ( t == 0 ){ printf("0"); retur.

直接输入字符串，然后逆序#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int revrse(char *arr, int len){ int i, k = 0; char ptr[len + 1]; for (i = len - 1; i >= 0; i--){ ptr[k] = arr[i];.

1.找一个y使f（x，y）=z，已知x，z，且由题目，z中的每一个字符一定都比x中的小。2.故如果z中有一个字符比x中的大，就一定找不到解，可以直接输出-1，结束程序了。3.如果没有，那么z本身也可以作为一个y的解，由于输出任意一个，所以那还找啥，直接输出z。注：全过程没有y的事儿，不用管y#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include&.

模拟 搜索 枚举 暴力 拆分#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int n, m, r, i, j, k, x, y, cnt = 0; scanf("%d %d %d", &n, &m, &r); int sign[105].

#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int n, i; scanf("%d\n", &n); int num[n]; for ( i = 0; i < n; i++) scanf("%d", &num[i]);.

康拖展开的函数：int Contor(char s[],int n){ int i, j; int ans=0; for(i=0;i<n;i++) { int smaller=0; for(j= i+1 ;j<n;j++) { if(s[i] > s[j])smaller++; } ans += smaller*fac[n-i-1]; } return ans+1;}AC代码：#include<stdio.h>const ..

自己写了快一个小时，数组越界了，只得了20分#include<stdio.h>#include<ctype.h>#include<string.h>#include<math.h>#include<stdlib.h>int main(){ int n, m, i, j, k, l, cnt = 0; scanf("%d %d %d", &n, &m, &k); //m是火把数，k是萤石数.