洛谷P1005 矩阵取数游戏 DP+高精类

#include<cstdio>
#include<cstring>

using namespace std;

inline int max(int x,int y){
    if(x>y)return x;else return y;
}

struct bignum{
    int c[10],l;
    
    void clear(){
        memset(c,0,sizeof(c));
        l=1;
    }
    
    bignum(){
        clear();
    }
    
    bignum operator =(int x){
        memset(c,0,sizeof(c));
        l=0;
        while(x){
            l++;
            c[l]=x%10000;
            x/=10000;
        }
        return *this;
    }
    
    bignum(int x){
        *this=x;
    }
    
    void print(){
        printf("%d",c[l]);
        for(int i=l-1;i>=1;i--){
            if(c[i]<10)printf("000");
            else if(c[i]<100)printf("00");
            else if(c[i]<1000)printf("0");
            printf("%d",c[i]);
        }
    }
};

bignum operator +(bignum a,bignum b){
    bignum ans;
    ans.l=max(a.l,b.l); 
    for(int i=1;i<=ans.l;i++)ans.c[i]=a.c[i]+b.c[i];
    for(int i=1;i<=ans.l;i++)
        if(ans.c[i]>=10000){
            ans.c[i]-=10000;
            ans.c[i+1]++;
        }
    if(ans.c[ans.l+1])ans.l++;
    return ans;
}

bignum operator *(bignum a,bignum b){
    bignum ans;
    for(int i=1;i<=a.l;i++)
        for(int j=1;j<=b.l;j++){
            ans.c[i+j-1]+=a.c[i]*b.c[j];
            ans.c[i+j]+=ans.c[i+j-1]/10000;
            ans.c[i+j-1]%=10000;
        }
    if(ans.c[a.l+b.l])ans.l=a.l+b.l;else ans.l=a.l+b.l-1;
    return ans;
}

bool operator <(bignum a,bignum b){
    if(a.l!=b.l)return a.l<b.l;
    for(int i=a.l;i>=1;i--)if(a.c[i]!=b.c[i])return a.c[i]<b.c[i];
    return 0;
}

bool operator >(bignum a,bignum b){
    return b<a;
}

bignum dfs(int,int);

int n,m;
int c[82];
bignum f[82][82];
bool pd[82][82];  //0表示未使用 
bignum ans;
bignum fz[82];

int main(){
    scanf("%d%d",&n,&m);
    fz[1]=2;
    for(int i=2;i<=m;i++)fz[i]=fz[i-1]*bignum(2);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)scanf("%d",&c[j]);
        memset(pd,0,sizeof(pd));
        ans=ans+dfs(1,m);
    }
    ans.print();
    printf("
");
    
    return 0;
}

bignum dfs(int l,int r){  //边界判断置于上一层,进入函数的l,r合法 
     if(pd[l][r])return f[l][r];
     pd[l][r]=1;
     if(l==r)return f[l][r]=bignum(c[l])*fz[m];
     bignum ans1=bignum(c[l])*fz[m-(r-l)]+dfs(l+1,r);
     bignum ans2=bignum(c[r])*fz[m-(r-l)]+dfs(l,r-1);
     if(ans2<ans1)return f[l][r]=ans1;
     else return f[l][r]=ans2;
}