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题目描述
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
输入
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
输出
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
示例输入
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
示例输出
Case 1: 14 1 4Case 2: 7 1 6
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int ls[100005];
int begin[100005];
int main()
{int t;scanf("%d",&t);for(int r = 1; r<=t; r++){int n;scanf("%d",&n);int i;for(i = 1; i <= n; i++)scanf("%d",&ls[i]);int max= 0, sum = 0;int top = 0, end = 1;max = ls[1];sum = ls[1];begin[++top] = 1;for(i = 2; i <= n; i++){sum += ls[i];if(sum >= max){end = i;max = sum;}else if(sum < 0){sum = 0;begin[++top] = i+1;}}printf("Case %d:\n",r);while(begin[top]>end){top--;}if(r < t){printf("%d %d %d\n",max,begin[top],end);printf("\n");}else{printf("%d %d %d\n",max,begin[top],end);}}return 0;
}
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