动态规划练习28:Maximum sum
2017-04-19 17:13
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题目简要:
描述Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
Your task is to calculate d(A).输入The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.输出Print exactly one line for each test case. The line should contain the integer d(A).样例输入
样例输出
提示In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
这道题就是求最大二子段的问题。
解题思路:
对于每一个数都求一下有这个数的前面的最大子段和,还有后面的最大字段和,最后加起来,放到数组里,求最大就好啦。
附代码:
#include<bits/stdc++.h>
using namespace std;
int a[50005],b[50005],c[50005];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
int sum=0;
b[0]=-100000;
c[n+1]=-1000000;
for(int i=1;i<=n;i++)
{
if(b[i-1]>sum+a[i])
b[i]=b[i-1];
else b[i]=sum+a[i];
sum=sum+a[i];
if(sum<0)
sum=0;
}
sum=0;
for(int i=n;i!=0;i--)
{
if(c[i+1]>sum+a[i])
c[i]=c[i+1];
else c[i]=sum+a[i];
sum=sum+a[i];
if(sum<0)
sum=0;
}
int answer=-1000000;
for(int i=1;i<=n;i++)
{
if(b[i]+c[i+1]>answer)
answer=b[i]+c[i+1];
}
cout<<answer<<endl;
}
return 0;
}
解题感受:
其实这个问题只要想到就好求啦,虽然WR了4次····
描述Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
t1 t2 d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n } i=s1 j=s2
Your task is to calculate d(A).输入The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.输出Print exactly one line for each test case. The line should contain the integer d(A).样例输入
1 10 1 -1 2 2 3 -3 4 -4 5 -5
样例输出
13
提示In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
这道题就是求最大二子段的问题。
解题思路:
对于每一个数都求一下有这个数的前面的最大子段和,还有后面的最大字段和,最后加起来,放到数组里,求最大就好啦。
附代码:
#include<bits/stdc++.h>
using namespace std;
int a[50005],b[50005],c[50005];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
int sum=0;
b[0]=-100000;
c[n+1]=-1000000;
for(int i=1;i<=n;i++)
{
if(b[i-1]>sum+a[i])
b[i]=b[i-1];
else b[i]=sum+a[i];
sum=sum+a[i];
if(sum<0)
sum=0;
}
sum=0;
for(int i=n;i!=0;i--)
{
if(c[i+1]>sum+a[i])
c[i]=c[i+1];
else c[i]=sum+a[i];
sum=sum+a[i];
if(sum<0)
sum=0;
}
int answer=-1000000;
for(int i=1;i<=n;i++)
{
if(b[i]+c[i+1]>answer)
answer=b[i]+c[i+1];
}
cout<<answer<<endl;
}
return 0;
}
解题感受:
其实这个问题只要想到就好求啦,虽然WR了4次····
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