题目:
题解:
新姿势了昂
yts中就是高精度的常见操作,比如c=a+b,你在结构体内部重载+的时候,只需要带进去一个b,a就是yts本身的a,啊总之就是特别好用
你要是想用max就定义max,如果你想对这些高精度数字排序就重载<
代码:
#include <cstdio>
#include<bits/stdc++.h>
#include <cstring>
#include <iostream>
#define LL long long
#define N 100000
using namespace std;
struct yts
{
int len,a[81];
yts en(){memset(a,0,sizeof(a));}
yts operator= (yts b)
{
len=b.len;
for (int i=b.len;i>=1;i--)
a[i]=b.a[i];
}
yts operator+ (yts b)
{
yts c;int p=0;
c.en();
c.len=max(len,b.len);
for (int i=1;i<=c.len;i++)
c.a[i]=a[i]+b.a[i]+p , p=c.a[i]/10 , c.a[i]%=10;
if (p) c.a[++c.len]=p;
return c;
}
yts operator* (yts cc)
{
yts c;
c.en();
c.len=len+cc.len;
for (int i=1;i<=len;i++)
for (int j=1;j<=cc.len;j++)
c.a[i+j-1]+=a[i]*cc.a[j] , c.a[i+j]+=c.a[i+j-1]/10 , c.a[i+j-1]%=10;
while (!c.a[c.len] && c.len>1)
c.len--;
return c;
}
}ans,ans1,f[81][81],a[81],mi[81];
yts max(yts a,yts b)
{
if (a.len>b.len) return a;
if (b.len>a.len) return b;
for (int i=a.len;i>0;i--)
{
if (a.a[i]>b.a[i]) return a;
if (b.a[i]>a.a[i]) return b;
}
return a;
}
void print(yts a)
{
for (int i=a.len;i>0;i--) printf("%d",a.a[i]);
printf("\n");
}
int main()
{
int n,m,i,j,k;
scanf("%d%d",&n,&m);
ans.len=1; ans.a[0]=0;
mi[0].a[1]=1; mi[0].len=1;
for (i=1;i<=m;i++)
mi[i]=mi[i-1]+mi[i-1];
for (i=1;i<=n;i++)
{
memset(f,0,sizeof(f));
for (j=1;j<=m;j++)
{
int x;
scanf("%d",&x);
a[j].len=0;
while (x)
{
a[j].a[++a[j].len]=x%10;
x/=10;
}
}
ans1.en();
for (j=0;j<=m;j++)
for (k=m+1;k>j;k--)
{
if (j!=0) f[j][k]=f[j-1][k]+mi[j+m-k+1]*a[j];
if (k!=m+1) f[j][k]=max(f[j][k],f[j][k+1]+mi[j+m-k+1]*a[k]);
}
ans1.len=1;
for (j=0;j<=m;j++) ans1=max(ans1,f[j][j+1]);
ans=ans+ans1;
}
print(ans);
}
上面仅保留一种方式,下面还是用我熟悉的方法写高精度
压6位是擦边哦,f[i][j]表示剩[i,j]的时候的最大价值
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int I=1e6;
int a[85];
struct bignum{int l,a[105];}mi[85],f[85][85],ans;
void clear(bignum &a){a.l=0; memset(a.a,0,sizeof(a.a));}
bool operator <(bignum a,bignum b)
{
if (a.l==b.l)
{
for (int i=a.l;i>=1;i--)
if (a.a[i]!=b.a[i]) return a.a[i]<b.a[i];
}
else return a.l<b.l;
}
bignum operator +(bignum a,bignum b)
{
bignum c;
clear(c);
int x=0;
int hh=max(a.l,b.l);
for (int i=1;i<=hh;i++)
{
c.a[i]=a.a[i]+b.a[i]+x;
x=c.a[i]/I;
c.a[i]%=I;
}
c.l=hh;
if (x) c.a[++c.l]=x;
return c;
}
bignum gcd(bignum b,int num)
{
int i,x=0;
bignum a;
clear(a);
for (i=1;i<=b.l;i++)
{
a.a[i]=b.a[i]*num+x;
x=a.a[i]/I;
a.a[i]%=I;
}
a.l=b.l;
while (x) a.a[++a.l]=x%I,x/=I;
return a;
}
int main()
{
int n,m,i,l,j;
scanf("%d%d",&n,&m);
mi[0].l=1; mi[0].a[1]=1;
for (i=1;i<=m;i++) mi[i]=gcd(mi[i-1],2);
for (l=1;l<=n;l++)
{
bignum maxx;
clear(maxx);
for (i=1;i<=m;i++) scanf("%d",&a[i]);
for (i=1;i<=m;i++)
for (j=m;j>=i;j--)
f[i][j]=max(f[i-1][j]+gcd(mi[m-j+i-1],a[i-1]),f[i][j+1]+gcd(mi[m-j+i-1],a[j+1]));
for (i=1;i<=m;i++) maxx=max(maxx,f[i][i]+gcd(mi[m],a[i]));
ans=ans+maxx;
}
printf("%d",ans.a[ans.l]);
for (i=ans.l-1;i>=1;i--)
{
printf("%d",ans.a[i]/100000);
printf("%d",ans.a[i]/10000%10);
printf("%d",ans.a[i]/1000%10);
printf("%d",ans.a[i]/100%10);
printf("%d",ans.a[i]/10%10);
printf("%d",ans.a[i]%10);
}
}