杭电 1003 max sum
2015-12-04 21:11
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 191685 Accepted Submission(s): 44647
Problem Description
Given a sequence a[1],a[2],a[3]......a
, 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
/*此题妙处在于输出开始位置和结束位子*/
#include<stdio.h> #include<string.h> using namespace std; int dp[1212121]; int a[121212]; int st[121212];//st数组是一个很巧妙的用处~ int main() { int t; int kase=0; int ok=0; scanf("%d",&t); while(t--) { if(ok)printf("\n"); ok++; memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); int n; scanf("%d",&n); int fushu=0; for(int i=0;i<n;i++) { scanf("%d",&a[i]); if(a[i]>0)fushu=1; st[i]=i+1; } //////////// if(fushu==0) { int output=-0x1f1f1f1f; int s; int e; for(int i=0;i<n;i++) { if(a[i]>output){output=a[i];s=i;e=i;} } printf("Case %d:\n",++kase); printf("%d %d %d\n",output,s+1,e+1); continue; } dp[0]=a[0]; int k=0; int maxn=dp[0]; int sto; for(int i=1;i<n;i++) { if(a[i]+dp[i-1]>=a[i]){dp[i]=dp[i-1]+a[i];st[i]=st[i-1];} else dp[i]=a[i]; if(dp[i]>maxn){maxn=dp[i];k=i;sto=st[i];} } //k+1是结尾0.0; printf("Case %d:\n",++kase); printf("%d %d %d\n",maxn,sto,k+1); } }
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