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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
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.
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
1 7 2 21 3 2
Source
2009 Greater New York Regional
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/* 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; }
HDU 3280 Equal Sum Partitions(二分查找)
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