关于联电(13)代码解决方案的搜索结果

1：杭电 2201

熊猫阿波的故事Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4141 Accepted Subm

https://www.u72.net/daima/vr41.html - 2024-07-15 02:15:16 - 代码库

2：杭电 1001

#include<stdio.h>int main(){int i;int n;int sum;while(scanf("%d",&n) != EOF){sum = 0;for(i = 1;i <= n; ++ i){sum += i; }print

https://www.u72.net/daima/vr51.html - 2024-08-23 17:36:28 - 代码库

3：杭电2061

水题：代码如下：#include<stdio.h>#include<string.h>int main(){ int n,m,i,j,count; double c,sc;//设为double型才给通过 double s,s1,s2; char

https://www.u72.net/daima/vdu4.html - 2024-07-14 22:19:04 - 代码库

4：杭电 2060

SnookerTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6707 Accepted Submission(

https://www.u72.net/daima/vku2.html - 2024-07-14 21:26:39 - 代码库

5：杭电 1715

大菲波数Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11143 Accepted Submissio

https://www.u72.net/daima/vffu.html - 2024-07-14 23:59:12 - 代码库

6：杭电 2056

RectanglesTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14340 Accepted Submiss

https://www.u72.net/daima/vhvn.html - 2024-07-14 20:35:36 - 代码库

7：杭电2023

1 #include <stdio.h> 2 double a[51][6]; 3 int main() 4 { 5 int n,m,i,j,t; 6 double s,x; 7 while(~scanf("%d%d",&n,&m)) 8

https://www.u72.net/daima/3uhe.html - 2024-09-03 00:22:39 - 代码库

8：杭电2046

#include<stdio.h>int sum(int n){ if(n>2) return sum(n-1)+sum(n-2); else return n;}int main(){ int n; while(scanf("%d",&n)!=EOF) { printf(

https://www.u72.net/daima/3zcs.html - 2024-07-20 21:07:49 - 代码库

9：杭电1008

1 #include <iostream> 2 #define MAX 100 3 using namespace std; 4 int main(){ 5 int n,no; 6 int a[100];a[0]=0; 7 while(ci

https://www.u72.net/daima/r2nz.html - 2024-08-19 02:04:24 - 代码库

10：杭电 2037

考察的是结构体的二次快排 #include<stdio.h>#include<stdlib.h>typedef struct Node{ int fir; int end;}Node;Node node[110];int cmp(const

https://www.u72.net/daima/u4xz.html - 2024-07-14 10:48:22 - 代码库

11：杭电2024

1 #include<stdio.h> 2 char s[51]; 3 int lenth(char s[]) 4 { 5 int i=0; 6 while(s[i]) 7 ++i; 8 return i; 9 }10

https://www.u72.net/daima/1vva.html - 2024-08-30 16:59:56 - 代码库

12：杭电2072

1 //C++STL入门，set的简单应用 2 #include<iostream> 3 #include<string> 4 #include<sstream> 5 #include<set> 6 using namespace std; 7 set

https://www.u72.net/daima/765f.html - 2024-09-10 17:22:55 - 代码库

13：杭电1096

一定记住输出格式，最后一个输出时，不再输出换行#include<stdio.h>int main(){ int a,b,n,m; while(scanf("%d",&n)!=EOF) { while(n--) { int sum=0;

https://www.u72.net/daima/741s.html - 2024-07-25 16:52:38 - 代码库

14：杭电1114

1 //完全背包，有待改良 2 #include<stdio.h> 3 #include<string.h> 4 typedef struct 5 { 6 int val,room; 7 } cus; 8 int an[10005]; 9

https://www.u72.net/daima/4c50.html - 2024-09-04 14:01:41 - 代码库

15：电商秒杀

秒杀思路：后台根据商品加入秒杀发布秒杀详细信息，生成静态页面，发布信息添加到库，前台查看秒杀信息，（我前台的概念是买家使用的系统为前台，公司内部使用

https://www.u72.net/daima/4m9u.html - 2024-07-22 20:34:17 - 代码库

16：杭电2058

#include<stdio.h> int main() { int i,j,sum=0; int n,m; while(scanf("%d%d",&n,&m)&&n!=0&&m!=0) { if(n>m)

https://www.u72.net/daima/7uv3.html - 2024-07-25 09:08:25 - 代码库

17：杭电2546

#include<iostream>#include<algorithm>#include<string.h>using namespace std;int max(int a,int b){ if(a>b) return a; else return b;}int main

https://www.u72.net/daima/70xh.html - 2024-07-25 13:02:16 - 代码库

18：杭电2042

1 #include<stdio.h> 2 int main() 3 { 4 int i,n,m; 5 while(scanf("%d",&n)!=EOF) 6 { 7 while(n--) 8 { 9

https://www.u72.net/daima/3721.html - 2024-07-21 15:48:15 - 代码库

19：杭电2043

1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 int m,c,i; 6 char s[51]; 7 while(scanf("%d",&m)!=EOF) 8 { 9

https://www.u72.net/daima/376x.html - 2024-07-21 15:56:19 - 代码库

20：杭电2040

#include<stdio.h>int main(){ int m,x,y,i,j; while(scanf("%d",&m)!=EOF) { while(m--) { int sum1=0,sum2=0;

https://www.u72.net/daima/3597.html - 2024-07-21 14:14:36 - 代码库