题目来源:赛马网(Max Sum)
Problem Description
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.
Input
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).
Output
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.
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
#include <stdio.h> #include <string.h> #define MAXLEN 100001 const int MIN = -1000; int num[MAXLEN]; int main() { int caseN; while (scanf("%d", &caseN) != EOF) { int cas = 0; while (caseN--) { int n; int tmp = 1; int s = 0, e = 0; int sumMax = MIN; num[0] = 0; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &num[i]); num[0] += num[i]; if (num[0] > sumMax) { sumMax = num[0]; s = tmp; e = i; } if (num[0] < 0 ) { num[0] = 0; tmp = i + 1; } } cas++; printf("Case %d:\n",cas); printf("%d %d %d\n", sumMax,s,e); if (caseN) printf("\n"); } } return 0; }
原文:http://www.cnblogs.com/tgycoder/p/5334313.html