Love丶伊卡洛斯

#include <stdio.h>#include <stdlib.h>#include <string.h>int main(){ int m,n,s; //转发的总量、小明决定的中奖间隔、以及第一位中奖者的序号 scanf("%d %d %d",&m,&n,&s); int i,j; ...

#include <iostream>#include <stdio.h>using namespace std;int main(){ int n; cin>>n; int a[10000]; int num[n]; for(int i=0;i<n;i++) { cin&gt...

#include <stdio.h>#include <stdlib.h>int main(){ int n,m,k; //学生人数/需要被查缴的物品种类数/个人物品数量 scanf("%d%d",&n,&m); int capture[m]; //需要被查缴的物品种类数 char name[n][4];...

 #include <stdio.h>#include <stdlib.h>int main(){ int n,m; scanf("%d%d",&n,&m); int food[m][n]; //各个月饼的销量 int sum[n]; //总销量 int back[n]; //备份 int i=0...

 #include <stdio.h>#include <stdlib.h>int main(){ int m; scanf("%d",&m); int k[m],K[m]; int i=0,n=0,j=0; for(i=0;i<m;i++) { scanf("%d",&k[...

#include <stdio.h>#include <stdlib.h>#include <math.h>int main(){ int M,X,Y; scanf("%d%d%d",&M,&X,&Y); int j,y; double b; //丙是个陷阱 int i=0; ...

#include <stdio.h>#include <stdlib.h>int main(){ int N; scanf("%d",&N); int i=1; int num=0; int box[N]; //存储不重复值 int count=0; //记录不重复值的个数 for(i=1;i&lt...

#include <stdio.h>#include <stdlib.h>int main(){ int A=0,B=0; scanf("%d%d",&A,&B); int C=0; C=A*B; int result[10]; int count=0; while(C!=0) { ...

#include <stdio.h>#include <stdlib.h>int main(){ int N; scanf("%d",&N); char line[20]={}; char key[N]; int i,j; for(i=0;i<=N;i++) { gets(lin...

#include <stdio.h>#include <stdlib.h>int main(){ int T=0,K=0,i=0; int n1=0,b=0,t=0,n2=0; scanf("%d%d",&T,&K); int x = T; //初始筹码 for(; i<K; i++) { ...

import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int N,M; N=sc.nextInt(); M=sc.nextInt(); int[] text=new int[100];

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N=sc.nextInt(); int[] num= {0,0,0,0}; int A=0,B=0; for(int

#include using namespace std;int main(){ int g1,s1,k1,g2,s2,k2; char x='.'; //cin不能写成cin>>".">>s1; 这样就能解决问题 int P,A,money; cin>>g1>>x>>s1>>x>>k1; cin>>g2>>x>>s2>>x>>k2;

#include <iostream>#include <stdio.h>using namespace std;int main(){ int n; cin>>n; int ID[n]; //先用int存储，之后在输出时补0 int x[n],y[n]; //x 和 y 是其打出的弹洞的平面坐标(x,y)...

#include #include using namespace std;int main(){ int n; string num[1001]; int seat1[1001],seat2[1001]; cin>>n; for(int i=0;i<n;i++) { cin>>num[i]>>seat1[i]>>seat2

#include #include #include #include using namespace std;int main(){ int N,M; cin>>N>>M; getchar(); int score[M]; int Cnum[4]; int Tnum[4]; char answer[1000]; string str1[M

//编译器编译时可能会有些小问题，不过可以通过#include #include #include #include using namespace std;int main(){ string str; getline(cin,str); int n=str.length(); char * cstr = new char [n+1]; st

#include using namespace std;int main(){ int N,n; char C; cin>>N>>C; if(N%2!=0) //判断四舍五入部分 n = N/2+1; else n = N/2; for(int j=0;j<N;j++) {

#include <iostream>#include<stdio.h>using namespace std;//用最朴实的方法，没错，我就是这么loint main(){ int B1=0,B2=0,C1=0,C2=0,J1=0,J2=0,n; cin>>n; int p=0; char a,b; for(...

#include using namespace std;int main(){ int N; cin>>N; double len[N-1]; for(int i=0;i<N;i++) { cin>>len[i]; } for(int i=0;i<N;i++) { for(int j=0;j

#include &lt;iostream&gt;using namespace std;int main(){ int n; cin&gt;&gt;n; int x[2]; x[0]=n%10; x[1]=(n/10)%10; x[2]=n/100; string change; string a=""; string b=""; string c=...

#include &lt;iostream&gt;using namespace std;int main(){ int n; cin&gt;&gt;n; int x=0; while(n!=1) { if(n%2==0){ n=n/2; }else{ n=(3*n+1)/2; } x++; } ...