poj 2549 Sumsets (枚举+二分)

Sumsets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10026		Accepted: 2743

Description


Given S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S.
Input

Several S, each consisting of a line containing an integer 1 <= n <= 1000 indicating the number of elements in S, followed by the elements of S, one per line. Each element of S is a distinct integer between -536870912 and +536870911 inclusive. The last line
of input contains 0.
Output

For each S, a single line containing d, or a single line containing "no solution".
Sample Input

5
2
3
5
7
12
5
2
16
64
256
1024
0

Sample Output

12
no solution

Source
Waterloo local 2001.06.02

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n,data[1005];

int main ()
{
while (scanf("%d",&n),n)
{
int ans,i;
bool flag=true;
for (i=0;i<n;i++)
scanf("%d",&data[i]);
sort(data,data+n);
int sz = unique(data,data+n)-data; //去重
n=sz;
for (int d=n-1;d>=0 && flag;d--) //枚举d
for (i=0;i<=n-4 && flag;i++) //注意因为有负数存在,需要扫完全部
{
int head=i+1,tail=n-1; //从首尾两个方向扫描
while (head<tail)
{
int tmp=data[i]+data[head]+data[tail];
if (tmp==data[d])
{
if (i==d || head==d || tail==d) break;
flag=false;
ans=tmp;
break;
}
else if (tmp>data[d]) tail--;
else head++;
}
}
if (flag)
printf("no solution\n");
else
printf("%d\n",ans);
}
return 0;
}