HDU 3280 Equal Sum Partitions（二分查找）

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本文介绍了一个经典的算法问题——等和划分。问题要求将一组正整数序列划分成若干个子集，使得每个子集内的元素之和相等，并探讨了如何通过编程找到这种划分方式中子集和的最小可能值。

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

Problem Description

An equal sum partition of a sequence of numbers is a grouping of the numbers (in the same order as the original sequence) in such a way that each group has the same sum. For example, the sequence:
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.

Input

The first line of input contains a single integer 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.

Output

For each data set, generate one line of output with the following values: The data set number as a decimal integer, a space, and the smallest sum for an equal sum partition of the sequence.

Sample Input

  
   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

Sample Output

Source

2009 Greater New York Regional

Recommend

/*
 题意:n个数，分成若干个集合，要求每个集合的数和相同，求集合最小值
 思路：枚举当前集合判断是否满足条件

*/
#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;
}