HDU 3280 Equal Sum Partitions(二分查找)
Equal Sum Partitions
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 551 Accepted Submission(s): 409
2 5 1 3 3 7
may be grouped as:
(2 5) (1 3 3) (7)
to yield an equal sum of 7.
Note: The partition that puts all the numbers in a single group is an equal sum partition with the sum equal to the sum of all the numbers in the sequence.
For this problem, you will write a program that takes as input a sequence of positive integers and returns the smallest sum for an equal sum partition of the sequence.
P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by a decimal integer
M, (1 ≤ M ≤ 10000), giving the total number of integers in the sequence. The remaining line(s) in the dataset consist of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than
10 values.
3
1 6
2 5 1 3 3 7
2 6
1 2 3 4 5 6
3 20
1 1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1
1 7
2 21
3 2
/*
题意:n个数,分成若干个集合,要求每一个集合的数和同样,求集合最小值
思路:枚举当前集合推断是否满足条件 */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std; typedef __int64 ll; #define N 10005
#define INF 0x3f3f3f3f int sum[N];
int n; bool fdd(ll temp)
{
int hh=0;
int pos=0;
while(pos!=n)
{
hh+=temp;
pos=upper_bound(sum+1,sum+n+1,hh)-(sum+1);
if(sum[pos]!=hh)
{
return false;
}
}
return true;
} int main()
{
int i,j,t,ca;
sum[0]=0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&ca,&n);
int x;
for(i=1;i<=n;i++)
{
scanf("%d",&x);
sum[i]=sum[i-1]+x;
} for(i=1;i<=n;i++)
if(fdd(sum[i])) break; printf("%d %d\n",ca,sum[i]);
}
return 0;
}
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