超级坑的水题!!!想了两天没一点思路,看了题解第一段话就做出来了

刚开始一直在想找到通项就是例如an*10^n+...+a0*10^0-an-...-a0>=s,然后从这个里面找到规律,结果走进死胡同了

Let's prove that if x is really big, then x + 1 is really big too.

证明很容易,我就不证了,直接上代码,二分答案就行了

还有一点要注意,因为是10E18范围,所以平时循环50次是不够用的,加到了70次就ac了,也可以用l<r-1这个来判断

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=,inf=; ll n,s;
bool ok(ll x)
{
ll t=x,p=x;
while(p){
t-=(p%);
p/=;
}
return t>=s;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie();
cin>>n>>s;
ll l=,r=n+;
for(int i=;i<;i++)
{
ll m=(l+r)/;
if(ok(m))r=m;
else l=m;
}
cout<<n-l<<endl;
return ;
}

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