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Max Sum

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 255311????Accepted Submission(s): 60666


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).
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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 #include<iostream> using namespace std; int main() {int N,n,a,first,last;cin>>N;for(int i=1;i<=N;i++){cin>>n;int max=-1000,sum=0,k=1;for(int j=1;j<=n;j++){cin>>a;sum=sum+a;if(sum>max)//如果當(dāng)前的最大值大于以前的最大值 更新 sum 與max 都是一種計(jì)算的結(jié)果和屬性相同{first=k;last=j;max=sum;}if(sum<0)//else{k=j+1;sum=0;}}cout<<"Case "<<i<<":"<<endl;cout<<max<<" "<<first<<" "<<last;if(i==N)cout<<endl;elsecout<<endl<<endl;}return 0; }
#include <stdio.h> int main() {int z,n,max,sum;int a,b,A,B,t;scanf("%d",&z);for(int k=1;k<=z;k++){scanf("%d",&n);sum = max = -1001;for(int i=1;i<=n;i++){scanf("%d",&t);if(sum+t < t)sum = t , a = b = i; //a、b記錄當(dāng)前連續(xù)子序列的起始、結(jié)束位置elsesum += t , ++b;if(max < sum)//附一篇另一種代碼，max = sum , A = a , B = b;}printf("Case %d:\n%d %d %d\n",k,max,A,B);if(k-z) puts("");}return 0; }

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