1007. Maximum Subsequence Sum (25)

题目url:http://pat.zju.edu.cn/contests/pat-a-practise/1007

reference url:http://blog.csdn.net/xtzmm1215/article/details/39010843

/*经典DP问题, 基于这样一个事实：保存一个最大字段和以及一个当前子段和, 
如果当前字段和大于当前最大字段和, 那么更新这个最大字段和,
如果当前字段和为负数的时候, 直接把当前字段和甚设置成0, */

#include <cstdio>
#include <stdio.h>
int a[10001];
int main(){
    int k;
    scanf("%d", &k);
    for (int i = 0; i < k; i++){
        scanf("%d", &a[i]);
    }
    int sum = 0, start = 0, end = k - 1, temp = 0, tempi = 0, tempj = 0;
    for (int i = 0; i < k; i++){
        if (temp >= 0){
            temp += a[i];
            tempj = i;
        }
        else{
            temp = a[i];
            tempi = i;
            tempj = i;
        }
        if (temp > sum || (temp == 0 && end == k -1)){
            sum = temp;
            start = tempi;
            end = tempj;
        }
    }
    printf("%d %d %d\n", sum, a[start], a[end]);
    return 0;
}

the time complexity is O(n)