51Nod 1049最大子段和 | 模板

 

 

Input示例

6
-2
11
-4
13
-5
-2

Output示例

20

1.最大子段和模板

#include "bits/stdc++.h"
using namespace std;
#define rep(i, s, n) for(int i=s;i<n;i++)
#define LL long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define E 2.71828
#define MOD 1000000007
#define N 50010
LL a[N];
LL b[N];
int main()
{
    LL max1 = 0;
    int n;
    while(~scanf("%d",&n))
    {
        bool f=0;
        rep(i,0,n){
            scanf("%lld",&a[i]);
        }
        memset(b,0,n+1);
        max1=-INF;
        rep(i,0,n)
        {
           if(b[i-1]>0)
           {
             b[i] = b[i-1]+a[i];
           }else{
             b[i] = a[i];
           }
           if(b[i]>max1)
             max1 = b[i];
        }
        if(max1<0)
            max1=0;
        cout<<max1<<endl;
    }
    return 0;
}

 2.

预处理：前缀和
last：上一个正数的位置
dp[i]表示这个子段最后一个是i的最大和
状态转移：
如果前一个是非负数，dp[i]=dp[i-1]+a[i]
否则，dp[i]=max(a[i],dp[last]+sum[i]-dp[last])
原理跟最大字段和一样

#include "bits/stdc++.h"
using namespace std;
#define rep(i, s, n) for(int i=s;i<n;i++)
#define LL long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define E 2.71828
#define MOD 1000000007
#define N 50010
int n;
long long a[N],dp[N],sum[N];
int main()
{
    while(~scanf("%d",&n)){
        int last=0;
        rep(i,1,n+1) cin>>a[i];
        rep(i,1,n+1) sum[i]=sum[i-1]+a[i];
        rep(i,1,n+1)
        {
            if(a[i-1]>=0)
                dp[i]=dp[i-1]+a[i],last=i;
            else
                dp[i]=max(a[i],dp[last]+sum[i]-sum[last]);
        }
        LL ans=0;
        rep(i,1,n+1)
           ans=max(ans,dp[i]);
        printf("%lld",ans);
    }
    return 0;
}