HDU 1003.Max Sum【最大连续子序列和】【8月14】

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

就是让你求一个数组中，连续最大的子序列和，以及起始位置，终止位置。之前做过子序列和，就是用数组存前多少项的和，需要i~~j的子序列和，就用f[j]-f[i-1]就可以了。最大子序列和，只需得到前n项和的最大值，以及最小值就可以了。但是在维护的过程中发现，直接做，最小值可能出现在最大值之后，导致WA。修改了多次也没有AC，不得不另找方法。

思路：跟上面改变的不多，前n项和为负数的时候，就从第n+1项继续寻找最大值。代码如下：

#include<cstdio>int main(){    int T,kase=1,n,x;    scanf("%d",&T);    while(T--){        int sum=0,maxsum=-9999,start=0,end=0,t=1;//start记录起点，end记录终点，t记录模拟起点        scanf("%d",&n);        for(int i=1;i<=n;i++){            scanf("%d",&x);            sum+=x;            if(sum>maxsum){                maxsum=sum;                start=t;                end=i;            }            if(sum<0){//前i项为负数，从第i+1项重新做                sum=0;                t=i+1;            }        }        if(kase>1) printf("\n");        printf("Case %d:\n%d %d %d\n",kase++,maxsum,start,end);    }    return 0;}