HDU 1003 Max Sum

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 187999 Accepted Submission(s): 43803

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

Author

Ignatius.L

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求连续递增最大子序列的和 ///特别sb的一个错误wa了一整版

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 200050#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;int dp[maxn],a[maxn],st,ed,maxx;int main(){    int loop,n,cnt=1;    rd(loop);    while(loop--){        rd(n);        FOR(i,1,n)rd(a[i]);MT(dp,0);        dp[1]=a[1];        FOR(i,2,n)            dp[i]=max(a[i],dp[i-1]+a[i]);        maxx=dp[1];ed=1;        FOR(i,1,n)if(dp[i]>maxx){ed=i;maxx=dp[i];}        int s=0;        for(int i=ed;i>0;i--){            s+=a[i];            if(s==maxx)st=i;        }        printf("Case %d:\n%d %d %d\n",cnt++,maxx,st,ed);////maxx打成了s  太傻逼了        if(loop)putchar('\n');    }    return 0;}/*25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5*/

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