http://acm.hdu.edu.cn/showproblem.php?pid=1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 202859 Accepted Submission(s): 47418
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
题意：求一个数列的最大子序列和
简单的动态规划，用dp[i]来记录i为终点的序列的最大子段和
如果dp[i-1]>0 dp[i] = dp[i-1] + a[i]
否则dp[i] = a[i]
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<string> 6 #include<cmath> 7 using namespace std; 8 int a[110000], dp[110000]; 9 int main(){ 10 int T, n, i, j, x, y, cas, xx, yy; 11 cin>>T; 12 for(cas=1; cas<=T; cas++){ 13 cin>>n; 14 for(i=1; i<=n; i++) 15 cin>>a[i]; 16 int mmax = -100000; 17 memset(dp, 0, sizeof(dp)); 18 dp[1] = a[1]; 19 x = 1, y = 1; 20 xx = 1, yy = 1; 21 for(i=1; i<=n; i++){ 22 if(dp[i-1]>=0){ 23 dp[i] = dp[i-1] + a[i]; 24 yy = i; 25 } 26 else{ 27 dp[i] = a[i]; 28 xx = i; 29 yy = i; 30 } 31 if(mmax <= dp[i]){ 32 mmax = dp[i]; 33 y = yy; 34 x = xx; 35 } 36 } 37 // for(i=1; i<=n; i++) 38 // cout<<dp[i]<<" "; 39 // cout<<endl; 40 printf("Case %d: ",cas); 41 printf("%d %d %d ",mmax,x,y); 42 if(cas!=T) 43 cout<<endl; 44 } 45 return 0; 46 }
还可以穷举
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 int a[110000]; 8 int main(){ 9 int T, n, i, j, sum, k, ans, cas, x, y; 10 cin>>T; 11 for(cas=1; cas<=T; cas++){ 12 cin>>n; 13 for(i=1; i<=n; i++) 14 cin>>a[i]; 15 sum = 0; 16 ans = -1000000; 17 k = 1; 18 for(i=1; i<=n; i++){ 19 sum += a[i]; 20 if(sum>ans){ 21 ans = sum; 22 x = k; 23 y = i; 24 } 25 if(sum<0){ 26 sum = 0; 27 k = i+1; 28 } 29 } 30 printf("Case %d: ",cas); 31 printf("%d %d %d ",ans, x, y); 32 //cout<<cas<<" "<<T<<endl; 33 if(cas!=T) 34 printf(" "); 35 } 36 return 0; 37 }
这里的sum就相当于dp[i]