输出:
For each S, a single line containing d, or a single line containing “no solution”.
样例
输入样例
5
2
3
5
7
12
5
2
16
64
256
1024
0
输出样例
12
no solution
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF -0x3f3f3f3f
using namespace std;
int a[1010];
int main()
{
int n;
int i,j;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int ans=INF;
for(i=n-1;i>=0;i--)
{
for(j=n-1;j>=0;j--)
{
if(i==j)
continue;
int sum=a[i]-a[j];
int l=0,r=j-1;
while(l<r)
{
if(a[l]+a[r]==sum)
{
ans=a[i];
break;
}
if(a[l]+a[r]>sum)
r--;
else
l++;
}
if(ans!=INF)
break;
}
if(ans!=INF)
break;
}
if(ans==INF)
printf("no solution\n");
else
printf("%d\n",ans);
}
return 0;
}
http://www.voidcn.com/article/p-dquihsug-bs.html
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