A.有理数
签到题：直接用floor函数就行了，详细看代码
#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define f first
#define s second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const double eps=1e-8;
const ll maxn =1e3+200;
const int N = 1e4+10;
const ll mod=1e9+7;
//define
//--solve
void solve() {
int i,j,tt=1;
cin>>tt;
while(tt--){
ll p,q;
ll tmp;
cin>>p>>q;
if(p%q==0){
tmp=floor(p*1.0/q)-1;
}
else tmp=floor(p*1.0/q);
cout<<tmp<<endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
#ifdef debug
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
cin.tie(0);
cout.tie(0);
solve();
/*
#ifdef debug
fclose(stdin);
fclose(stdout);
system("out.txt");
#endif
*/
return 0;
}
B。硬币，实际上50，10都是有5的倍数，所以可以直接吧50，10元的看成是数个5元构成硬币，那问题就容易了，假设一个物品要v元，remain=v%5，就为剩下的还要给少个硬币才能凑够买一个v元物品，然后k=5-remain，这这个k其实就是找回的1元硬币数，详细看代码
#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define f first
#define s second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const double eps=1e-8;
const ll maxn =1e3+200;
const int N = 1e4+10;
const ll mod=1e9+7;
//define
ll c[maxn];
//
ll gcd(ll a,ll b) {
return b==0?a:gcd(b,a%b);
}
//--solve
void solve() {
int i,j,tt=1;
cin>>tt;
ll c1,c2,c4,c3,v;
while(tt--){
ll ans=0ll,sum=0;
cin>>c1>>c2>>c3>>c4>>v;
sum=c2*5+c3*10+c4*50;
ll remain=v%5;
ll k=0ll;
if(remain){
k=5-remain;
v+=k;
}
ans=k*(sum/v)+c1;
cout<<ans<<endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
#ifdef debug
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
cin.tie(0);
cout.tie(0);
solve();
/*
#ifdef debug
fclose(stdin);
fclose(stdout);
system("out.txt");
#endif
*/
return 0;
}
