http://acm.hdu.edu.cn/showproblem.php?pid=3280
用了简单的枚举。
Equal Sum Partitions
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 453 Accepted Submission(s): 337
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
1 7
2 21
3 2
#include<iostream> #include<cstring> #include<cstdio> using namespace std; int a[10005]; int main() { int i,j,t,n,m,sum,cursum,flag ,ans; scanf("%d",&t); while(t--) { flag=0; memset(a,0,sizeof(a)); scanf("%d%d",&n,&m); for(i=0;i<m;i++) scanf("%d",&a[i]); for(i=0;i<m;i++) { sum=0; for(j=0;j<=i;j++) sum+=a[j]; cursum=0; while(j<m) { cursum+=a[j]; if(cursum>sum) break; else if(cursum==sum) { j++; if(j==m) { printf("%d %d\n",n,sum); flag=1; } cursum=0; } else j++; if(flag) break; } if(flag) break; } if(i==m) printf("%d %d\n",n,sum); } return 0; } /* 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 */
HDU-3280 Equal Sum Partitions