考试---订错2--1

dp  太差了,爆零 首先是乱蒙 if((dp[i][k]+dp[k+1][j]+(a[i][k]*a[k+1][j])%100)<dp[i][j])  

                           其次 尝试用i,j,k做区间,应该用k代表区间长度

                            最后初始化不对  

#include<iostream>
#include<string.h>
#include<math.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
int dp[105][105];
int a[105][105];
int t;
int n;
void clean()
{
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));    
}
int main()
{
//freopen("wish.in","r",stdin);
//freopen("wish.out","w",stdout);
int i,j,k,l;
cin>>t;
for(l=1;l<=t;l++)
{
clean();
cin>>n;
for(i=1;i<=n;i++)
{ 
cin>>a[i][i];    
} 
if(n==2) {cout<<a[1][1]*a[2][2]<<endl;continue;}
for(i=1;i<n;i++)
{
for(j=i+1;j<=n;j++)
{
for(k=i;k<=j;k++)
{
a[i][j]=a[k][k]+a[i][j];
//a[i][j]=a[i][j]%100;
}    
}    
}
dp[1][2]=a[1][1]*a[2][2];
for(k=n-1;k>=1;k--)
for(i=1;i<=k;i++)
for(j=k+1;j<=n;j++)
{ 
if((dp[i][k]+dp[k+1][j]+(a[i][k]*a[k+1][j])%100)>dp[i][j])
{ 
//cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
dp[i][j]=dp[i][k]+dp[k+1][j]+a[i][k]*a[k+1][j];
a[i][j]=a[i][k]+a[k+1][j];
}
}
cout<<dp[1][n]<<endl;    

}


return 0;
}

与我的方法较接近的方法

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,cnt,t;
ll dp[109][109];
ll then[109][109],a[109];
int main()
{
    freopen("wish.in","r",stdin);
    freopen("wish.out","w",stdout);
    cin>>t;
    while(t--)//要初始化!!! 
    {
        memset(a,0,sizeof(a));    
        memset(then,0,sizeof(then));
        cin>>n;
        for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++)
        {
            dp[i][j]=99999999999999999;
        }
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            then[i][i]=a[i];
        }
        for(int i=1;i<n;i++)
        {
            dp[i][i+1]=a[i]*a[i+1];//此处需要注意
            then[i][i+1]=(a[i]+a[i+1])%100;
        }
        for(int i=1;i<=n;i++)
        dp[i][i]=0;//从i到自己不用费力
        for(int len=3;len<=n;len++)//区间长度 
        {
            for(int i=1;i<=n;i++)//区间起点
            {
                int j=i+len-1;
                if(j>n)break;
                for(int k=i;k<j;k++)//k前面的乘起来再乘后面的
                {
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+then[i][k]*then[k+1][j]);//注意此处不能mod
                    then[i][j]=(then[i][k]+then[k+1][j])%100;//注意此处必须mod,因为g下面的已经mod了m(i,j),此预处理只有m(i,i+1);
                }
            } 
            
        }
        cout<<dp[1][n]<<endl;
    }
    return 0;
}

更简单的方法---合并石子

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define ll long long
#define mod 100
#define cl(x) memset(x,0,sizeof x)
#define pl(x) memset(x,9999999,sizeof x)
using namespace std;
const int maxn=1000+5;
ll f[maxn][maxn],a[maxn],m[maxn][maxn];
ll n,t;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int main()
{
    freopen("wish.in","r",stdin);
    freopen("wish.out","w",stdout);
    t=read();
    while(t--)
    {
        pl(f);cl(m);cl(a);
        n=read();
        for(int i=1;i<=n;i++) 
        {
            int k=read();a[i]=k+a[i-1];f[i][i]=0;
        }
        for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++)
        m[i][j]=(a[j]-a[i-1])%mod;//即我a【i]【j】;
        for(int i=2;i<=n;i++)//区间大小
        for(int j=1;j<=n-i+1;j++)
        {
            int k=j+i-1;
            for(int h=j;h<k;h++)
            f[j][k]=min(f[j][k],f[j][h]+f[h+1][k]+m[j][h]*m[h+1][k]);
        }
        cout<<f[1][n]<<endl;
    }
    return 0;
}

 订正

#include<iostream>
#include<string.h>
#include<math.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
int dp[105][105];
int m[105][105];
int a[105];
int t;
int n;
void clean()
{
    memset(dp,0x7f,sizeof(dp));
    memset(a,0,sizeof(a));        
}
int main()
{
    //freopen("wish.in","r",stdin);
    //freopen("wish.out","w",stdout);
    int i,j,k,l;
    cin>>t;
    for(l=1;l<=t;l++)
    {
      clean();
      cin>>n;
      for(i=1;i<=n;i++)
      {     
            cin>>a[i];
            m[i][i]=a[i];
            dp[i][i]=0;
      } 
      for(i=1;i<=n-1;i++)
      {
                dp[i][i+1]=a[i]*a[i+1];
                m[i][i+1]=(a[i]+a[i+1])%100;
      }
      for(i=2;i<=n;i++)//cnt
           for(j=1;j<=n-i+1;j++)//strat
           {
                int h=i+j-1;
                for(k=j;k<h;k++)
                        {
                         dp[j][h]=min(dp[j][h],dp[j][k]+dp[k+1][h]+m[j][k]*m[k+1][h]);
                          m[j][h]=(m[j][k]+m[k+1][h])%100;
                         
                        }
         }
      cout<<dp[1][n]<<endl;           
          
    }
    
    
    return 0;
}
原文地址:https://www.cnblogs.com/voldemorte/p/7411810.html