动态规划: HDU1003Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 251171    Accepted Submission(s): 59503

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

 

Author

Ignatius.L

Problem : 1003 ( Max Sum )     Judge Status : Accepted

RunId : 21239601    Language : G++    Author : hnustwanghe

Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int INF = -(1<<30);

int main(){
int T,n,cnt = 0;
scanf("%d",&T);
while(T--){
int ans ,start=1,tail,tmp = 0, x,ts;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&x);
if(i==1){
ans = tmp = x;
start = tail = ts = 1;
}
else{
if(x > x + tmp){
tmp = x;
ts = i;
}
else tmp = tmp + x;
}
if(tmp > ans ){
ans = tmp;
start = ts,tail = i;
}
}
printf("Case %d:\n%d %d %d\n",++cnt,ans,start,tail);
if(T) printf("\n");
}
}

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int INF = -(1<<30);

int main(){
    int T,n,cnt = 0;
    scanf("%d",&T);
    while(T--){
        int ans ,start=1,tail,tmp = 0, x,ts;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(i==1){
                ans = tmp = x;
                start = tail = ts = 1;
            }
            else{
                if(x > x + tmp){
                    tmp = x;
                    ts = i;
                }
                else tmp = tmp + x;
            }
            if(tmp > ans ){
                ans = tmp;
                start = ts,tail = i;
            }
        }
        printf("Case %d:\n%d %d %d\n",++cnt,ans,start,tail);
        if(T) printf("\n");
    }
}