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D_Dp
<span style="color:#000099;">/* D - 简单dp 例题 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status 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 By Grant Yuan 2014.7.16 */ #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> using namespace std; int t,n; int a[100005]; int dp[100005]; int ct=1; int main() { cin>>t; while(t--){ cin>>n; for(int i=1;i<=n;i++) cin>>a[i]; memset(dp,0,sizeof(dp)); int max=-999999; int l1,r1,l=1,r=1; for(int i=1;i<=n;i++) { if(dp[i-1]>=0) dp[i]=dp[i-1]+a[i],r=i; else dp[i]=a[i],l=i; if(dp[i]>max) max=dp[i],l1=l,r1=r; } if(r1<l1) r1=l1; printf("Case %d:\n",ct++); printf("%d %d %d\n",max,l1,r1); if(t) cout<<endl; } } </span>
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