Codeforces 1409D. Decrease the Sum of Digits

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
 
ll sumdigit(ll n){
	ll temp=0;
	for(;n;n/=10)
		temp+=n%10;
	return temp;
}
 
ll solve(ll n,ll s){
	if(sumdigit(n)<=s)
		return 0;
	if(n%10==0)
		return solve(n/10,s)*10;
	return solve(n+10-n%10,s)+10-n%10;
}
 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	ll n,s;
	cin>>t;
	for(int i=0;i<t;++i){
		cin>>n>>s;
		cout<<solve(n,s)<<endl;
	}
	return 0;
}

思路:輸入的大數n每個位數相加必須<=給定的s，則每次取尾數出來算要加的值(10-n%10)然後加上之前的值