HDOJ 1003-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
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

思路：套模板 用beg来记录最大和子串的开始位置 然后通过开始位置求末位置

#include"stdio.h"#include"math.h"#include"stdlib.h"const int maxn=10e5+100;int  ai[maxn];int mmax(int a,int b){    return (a>b)?a:b;}int maxsum(int x,int y,int &beg){    int a,b;    if(x==y)    {        beg=x;        return ai[x];    }    int mid=(x+y)/2;    int asum=maxsum(x,mid,a),bsum=maxsum(mid+1,y,b);    int maxs=mmax(asum,bsum);    int begs=(asum>bsum)?a:b;    int l=-10e8,r=-10e8,vis=0,kais;    for(int i=mid;i>=x;i--)    {        vis+=ai[i];        if(vis>l)        {            l=vis;            kais=i;        }    }    vis=0;    for(int i=mid+1;i<=y;i++)    {        vis+=ai[i];        r=mmax(r,vis);    }    beg=(maxs>(l+r))?begs:kais;    return mmax(maxs,l+r);}int main(){    int T,n;    int beg;    scanf("%d",&T);    int s=T;    while(T--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&ai[i]);        }        int big=maxsum(0,n-1,beg);        int sum=0,endd;        for(int i=beg;i<n;i++)        {            sum+=ai[i];            if(sum==big)            {                endd=i;                break;            }        }            printf("Case %d:\n",s-T);        printf("%d %d %d\n",big,beg+1,endd+1);        if(T)            printf("\n");    }    return 0;}