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2#include<stdio.h> int a[22][22]; int main() { int m,n,i,j; while(scanf("%d%d",&n,&m)==2) { a[1][1]=1; for(i=1;i<=n;i++) { i
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0hungry sentence code force problem 较一一判断生成素数省时 #include <iostream> #include <cstdio> #include <cstring> using n
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0#include<stdio.h> #include <string.h> #define MAXN 10000 int lastdigit(char* buf) { const int mod[20]={1,1,2,6,4,2,2,4,2,8,4,4,8
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0#include<stdio.h> int a[100000],b[100000]; int main() { int i,j,k,n,t,T,x=1; long s,smax; scanf("%d",&T); while(T--) { scanf("%d",
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0其实想到方法很简单 #include<stdio.h> #include<math.h> int main() { int n,a[1005],b[1005]; while(scanf("%d",&n)==1&&n
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0#include<stdio.h> #include<string.h> int a[1005]; char b[1005]; int main() { int T,n,k,i,tmp; scanf("%d",&T); while(T--) { s
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0#include<stdio.h> #include<stdlib.h> int main() { int i,j,n,t,x,max,lmax,a[1010],b[1010]; while(scanf("%d",&n)==1) { for(i=1
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0http://blog.csdn.net/a342374071/article/details/6694452 http://blog.csdn.net/yysdsyl/article/details/4226630
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0#include<stdio.h> int a[55]; int main() { int i,n,m,T; scanf("%d",&T); while(T--) { scanf("%d%d",&m,&n); a[1]=0;a[2]=1;a[3
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0#include<stdio.h> long a[50]={0,1,2}; int main() { int n,i; while(scanf("%d",&n)==1) { for(i=3;i<=n;i++) { a[i]=a[i-1]+a[i-2];
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0#include<stdio.h> long a[3005]={0,1}; int main() { int n,i,T; scanf("%d",&T); while(T--) { scanf("%d",&n); long maxn=0; for(i=
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2#include <cstdio>#include <algorithm>using namespace std;int n, m, a[1000010];bool cmp(int a, int b){ return a > b;}int main(
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0Problem C: 指针： 有n个整数，使其前面各数顺序向后移m个位置，最后m个数变成最前面m个数 Time Limit: 1 Sec Memory Limit: 64 MB Submit: 1369 Solv
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0输入输出和常用库函数 v掌握scanf与printf函数的定义和使用（#include <stdio.h>）–int：scanf("%d",&n);printf("%d", n);–float:scanf("
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0#include<stdio.h> #include<math.h> int main() { int n; while(scanf("%d",&n)!=EOF) { int sum=0,i,j=0; if(n==0)j=1; if(n<0)
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1#include<stdio.h> #include<math.h> int main() { int origin_num ; int counter ; int sum , single ,ed=1 ; float order; int fix=1 ;
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0#include<stdio.h> /*也可以在3883基础上修改，输出下标号不等于index的所有元素*/ int main() { int i, j, n, m, a[20]; while (scanf("%d", &
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0水
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