TOJ1398正方形的编成 或者 POJ2362

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int len[20],sum,M;
bool flag,f[20];//flag标记是否找到
void dfs(int d,int stick,int nowlen)
{
    //i：正在处理的小棍，stick：已经拼好的边框数，nowlen：当前正在拼的框的长度
    int j;
    if(flag||stick>2)
	{
        if(stick>2)
			flag=1;//已经拼好3个框，第四个不用拼了
        return ;
    }
    else
	{
        for(j=d;j<M;j++)
		{
            if((!f[j])&&nowlen+len[j]<=sum)
			{
                f[j]=1;//标记
                if(nowlen+len[j]==sum)
					dfs(0,stick+1,0);//拼好了一个，继续拼下一个
                else
					dfs(j+1,stick,nowlen+len[j]);//继续拼当前的
                f[j]=0;//恢复标记
            }
        }
    }
}
int main()
{
    int ca,i,j;
    scanf("%d",&ca);
    while(ca--)
	{
        scanf("%d",&M);
        for(i=sum=0;i<M;i++)
		{
            scanf("%d",&len[i]);
            sum+=len[i];
        }
        if(sum%4!=0)
		{
			printf("no\n");
			continue;
		}
        sum/=4;
        for(int i=0;i<M;i++)
			for(int j=i+1;j<M;j++)
			{
				if(len[i]<len[j])
				{
					int t=len[i];
					len[i]=len[j];
					len[j]=t;
				}
			}
        if(len[0]>sum)
		{
			printf("no\n");
			continue;
		}
        flag=0;
        dfs(0,0,0);
        printf("%s\n",flag?"yes":"no");
    }
    //system("pause");
    return 0;
}

　　

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

The first line of input contains N, the number of test cases.  Each test case begins with an integer 4 ≤ M ≤ 20, the number of sticks.  M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

Sample Input

3

#include <stdio.h>
int a[20];
int sum=0;
int b[20]={0};
int n;
int flag=0;
void fun(int step,int num,int toile)
{
	if(num>2)
	{
		flag=1;
	}

	for(int i=step;i<n;i++)
	{
		if((!b[i])&&(toile+a[i]<=sum))
		{
			b[i]=1;
			if(toile+a[i]==sum)
			{
				fun(0,num+1,0);
			}
			else
			{
				fun(i+1,num,toile+a[i]);
			}

			b[i]=0;

		}
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		sum=sum+a[i];
	}

	if(sum%4==0)
	{
		sum=sum/4;
		fun(0,0,0);
	}

	else
	{
		printf("NO");
	}
	if(flag==1)
		printf("YES");
	return 1;
}

#include <stdio.h> int a[20]; int sum=0; int b[20]={0}; int n; int flag=0; void fun(int step,int num,int toile) { if(num>2) { flag=1; } for(int i=step;i<n;i++) { if((!b[i])&&(toile+a[i]<=sum)) { b[i]=1; if(toile+a[i]==sum) { fun(0,num+1,0); } else { fun(i+1,num,toile+a[i]); } b[i]=0; } } } int main() { scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); sum=sum+a[i]; } //先排序 for(int i=0;i<n;i++) for(int j=i+1;j<n;j++) { if(a[i]>a[j]) { int t=a[i]; a[i]=a[j]; a[j]=t; } } if(sum%4==0) { sum=sum/4; fun(0,0,0); } else { printf("NO"); } if(flag==1) printf("YES"); return 1; }

4 1 1 1 1 5 10 20 30 40 50 8 1 7 2 6 4 4 3 5

Output for Sample Input

yes
no
yes