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 testcases. 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 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 -5Sample Output
Case 1: 14 1 4 Case 2: 7 1 6题意：求最大子序列和
思路：设置变量max记录最大值，设置下标beg和end记录组成max的元素首与尾。遍历数组，累加求和sum ；如果sum<0时，以当前元素为首元素(t_beg)从新开始，否则继续累加；仅仅当sum>max时，更新beg,end,max。
#include <algorithm> #include <cstdio> #include <cstring> #include <iostream> #include <vector> #define maxn 100001 using namespace std; int a[maxn]; int main() { memset(a, 0, sizeof(a)); int T, Case = 0; cin >> T; while (T--) { int n, i; cin >> n; for (i = 0; i < n; i++) { int b; cin >> b; a[i] = b; } int max = -1001, end = 0, sum = 0, beg = 0, tbeg = 0; for (i = 0; i < n; i++) { sum += a[i]; if (sum > max) { max = sum; end = i; beg = tbeg; } if (sum < 0) { sum = 0; tbeg = i + 1; } } printf("Case %d:\n", ++Case); printf("%d %d %d\n", max, beg + 1, end + 1); if (T) { printf("\n"); } } system("pause"); }转载于:https://www.cnblogs.com/ruoh3kou/p/9893461.html
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