BNUOJ 5227 Max Sum

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Max SumTime Limit: 1000msMemory Limit: 32768KB This problem will be judged on HDU. Original ID: 1003
64-bit integer IO format:  %I64d      Java class name:  Main  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.
InputThe 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).
 OutputFor 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


解題：dp入門題！弱菜的成長之路啊！

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 int dp[100010],num[100010];12 int main(){13     int kase,i,index,ans,n,k = 1;14     scanf("%d",&kase);15     while(kase--){16         scanf("%d",&n);17         for(i = 1; i <= n; i++)18             scanf("%d",dp+i);19         num[1] = 0;20         ans = dp[index = 1];21         for(i = 2; i <= n; i++){22             if(dp[i] <= dp[i-1]+dp[i]){23                 dp[i] = dp[i-1]+dp[i];24                 num[i] = num[i-1]+1;25             }else num[i] = 0;26             if(ans < dp[i]) ans = dp[index = i];27         }28         printf("Case %d:\n",k++);29         printf("%d %d %d\n",ans,index-num[index],index);30         if(kase) putchar(‘\n‘);31     }32     return 0;33 }

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