HDU 1003:Max Sum
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 174588 Accepted Submission(s): 40639
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
求和与end都没什么好说的。求start是亮点,一开始判断了很久,后来还是那个想法,对于序列或是字符串正着想想不出来就逆过来想,从左到右end容易求,那从右至左的话start就容易求了。
代码:
#include <iostream> #include <string> #include <cstring> #include <algorithm> #include <cmath> #pragma warning(disable:4996) using namespace std; int value[100005]; int dp[100005]; int main() { //freopen("input.txt","r",stdin); //freopen("out.txt","w",stdout); int Test,num,i,j,ans,start,end; value[0]=0; cin>>Test; for(j=1;j<=Test;j++) { ans = -1005; memset(dp,0,sizeof(dp)); cin>>num; for(i=1;i<=num;i++) { cin>>value[i]; dp[i]=max(dp[i-1]+value[i],value[i]); if(dp[i]>ans) { ans=dp[i]; end=i; } } ans=-1005; memset(dp,0,sizeof(dp)); for(i=num;i>=1;i--) { dp[i]=max(dp[i+1]+value[i],value[i]); if(dp[i]>=ans) { ans=dp[i]; start=i; } } cout<<"Case "<<j<<":"<<endl; cout<<ans<<" "<<start<<" "<<end<<endl; if(j!=Test) cout<<endl; } return 0; }
后来看discuss里其他人的思路,觉得从end开始往回倒,对每一个值都加起来,什么时候等于求出来的最大和了,什么时候就是start了,觉得这样做也很好。
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