最大连续子序列和

最大连续子序列和

因为最大 连续子序列和只可能是以位置0～n-1中某个位置结尾。当遍历到第i个元素时，判断在它前面的连续子序列和是否大于0，如果大于0，则以位置i结尾的最大连续子序列和为元素i和前门的连续子序列和相加；否则，则以位置i结尾的最大连续子序列和为元素i。

例题

相关题目练习：

hdu 1003：http://acm.hdu.edu.cn/showproblem.php?pid=1003

代码

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
int main()
{

    int n,k;
    int pos1 = 1, pos2 = 1, x = 1;
    int maxx, now;
    scanf("%d", &n);
    for (int j = 1; j <=n; j++)
    {
        int temp;
        scanf("%d", &k);
        scanf("%d", &temp);
        pos1 = 1, pos2 = 1, x = 1;
        maxx = now = temp;
        for (int i = 2; i <= k; i++)
        {
            scanf("%d", &temp);    
            if (now + temp < temp) { now = temp; x = i; }
            else now += temp;
            if (now > maxx)
            {
                maxx = now;
                pos1 = x; pos2 = i;
            }
        }
            printf("Case %d:
", j);
            printf("%d %d %d
", maxx, pos1, pos2);
            if (j != n)printf("
");
        
    }
}

hdu 1231：http://acm.hdu.edu.cn/showproblem.php?pid=1231

 代码

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
int dp[10003];
int main()
{

    int n;
    int pos1 = 1, pos2 = 1, x = 1;
    int maxx, now;
    while (1)
    {
        scanf("%d", &n);
        if (n == 0)break;
        memset(dp, 0, sizeof(dp));
            int temp;
            scanf("%d", &dp[1]);
            pos1 = 1, pos2 = 1, x = 1;
            maxx = now = dp[1];
            for (int i = 2; i <= n; i++)
            {
                scanf("%d", &dp[i]);
            }
            for (int i = 2; i <= n; i++)
            {
                if (now + dp[i] < dp[i]) { now = dp[i]; x = i; }
                else now += dp[i];
                if (now > maxx)
                {
                    maxx = now;
                    pos1 = x; pos2 = i;
                }
            }
            if(maxx<0)    printf("%d %d %d
", 0, dp[1], dp[n]);
            else printf("%d %d %d
", maxx, dp[pos1], dp[pos2]);
        }
}

参考：https://blog.csdn.net/sgbfblog/article/details/8032464