本地测试良好
提交出现CE (已尝试所有版本)
luogu IDE :
Nothing is compiled: OUTPUT exceeds.
求大佬解答
Code
#include<bits/stdc++.h>
using namespace std;
template<size_t _K=4>
class BigNum{
private:
int K=4;
int base=pow(10,K);
void __change_size(bool _Tp=0){
if(!_Tp){
while(c[c[0]+1])
c[0]++;
}
}
public:
int c[1005]={};
BigNum()=default;
BigNum(int x){
c[0]=0;
while(x){
c[++c[0]]=x%base;
x/=base;
}
}
BigNum(string x){
reverse(x.begin(),x.end());
int k=1;
c[0]=(x.size()+K-1)/K;
int t=0;
for(int i=0;i<x.size();i++,k*=10){
if(i%K==0)
k=1,t++;
c[t]+=(x[i]-'0')*k;
}
}
int size()const{
return c[0];
}
bool empty()const{
return c[0]==0;
}
BigNum operator+(const BigNum& A){
BigNum tmp(0);
auto &a=tmp.c;
auto &b=A.c;
a[0]=c[0];
for(int i=1;i<=max(b[0],c[0]);i++){
a[i]+=b[i]+c[i];
if(a[i]>=base)
a[i]-=base,a[i+1]++;
}
tmp.__change_size();
return tmp;
}
BigNum operator*(int &A){
if(A==0) return BigNum(0);
if(this->empty()) return BigNum(0);
BigNum tmp(0);
auto& a=tmp.c;
for(int i=1;i<=c[0]+5;i++){
a[i]+=c[i]*A;
a[i+1]+=a[i]/base;
a[i]%=base;
}
tmp.__change_size();
return tmp;
}
BigNum operator*(int &&A){
if(A==0) return BigNum(0);
if(this->empty()) return BigNum(0);
BigNum tmp(0);
auto& a=tmp.c;
for(int i=1;i<=c[0]+5;i++){
a[i]+=c[i]*A;
a[i+1]+=a[i]/base;
a[i]%=base;
}
tmp.__change_size();
return tmp;
}
void print(){
if(this->empty())
putchar('0');
else{
printf("%d",c[c[0]]);
for(int i=c[0]-1;i>0;i--)
printf("%0*d",K,c[i]);
}
}
};
typedef BigNum<4> BigInt;
template<size_t _K>
bool operator < (const BigNum<_K>& A,const BigNum<_K>& B){
if(B.size()!=A.size())
return (A.size())<(B.size());
for(int i=A.size();i>=1;i--)
if(A.c[i]!=B.c[i])
return A.c[i]<B.c[i];
return 0;
}
int a[85][85];
BigInt f[85];
BigInt dp[85][85];
int main(){
int n,m;
cin>>n>>m;
f[0]=1;
for(int i=1;i<=m;i++)
f[i]=f[i-1]*2;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
cin>>a[i][j];
BigInt ans=0;
for(int i=1;i<=n;i++){
for(int l=1;l<=m;l++)
for(int r=1;r<=m;r++)
dp[l][r]=0;
for(int len=1;len<=m;len++){
for(int l=1;l<=m&&l+len-1<=m;l++){
int r=l+len-1;
BigInt k=f[m-(r-l)];
dp[l][r]=max(dp[l+1][r]+k*a[i][l],dp[l][r-1]+k*a[i][r]);
}
}
ans=ans+dp[1][m];
}
ans.print();
}