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链接:
http://acm.hdu.edu.cn/showproblem.php?pid=1003
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 180817 Accepted Submission(s): 42261
5 6 -1 5 4 -7
6 5 10 14 7
Case 1:
14 1 4
7 0 6 -1 1 -6 7 -5
0 6 5 6 0 7 2
Case 2:
7 1 6
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #include <queue> using namespace std; #define N 110000 #define INF 0xffffff int sum[N]; int main() { int t, iCase=1; scanf("%d", &t); while(t--) { int i, n, Max=-INF, Min=INF, a, L, Left, Right; scanf("%d", &n); for(i=1; i<=n; i++) { scanf("%d", &a); sum[i] = sum[i-1]+a; }
/// Left = Right = 1; for(i=1; i<=n; i++) {
///找出最小的sum[i] if(sum[i-1]<Min) { Min = sum[i-1]; L = i; } ///找出最大的Max if(sum[i]-Min>Max) { Max = sum[i]-Min; Left = L; Right = i; } } printf("Case %d:\n", iCase++); printf("%d %d %d\n", Max, Left, Right); if(t) printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/YY56/p/4779905.html