#include<iostream>
using namespace std;
int main() {
    int T;
    cin >> T;
    while (T--) {
        int Xa, Ya, Xb, Yb, n;
        cin >> Xa >> Xb >> Ya >> Yb;
        if (Yb % Xa == 0) {
            n = Yb / Xa;
        }
        else
            n = Yb / Xa+1;
        if (n * Xb >= Ya) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}

#include <stdio.h>

bool ai(long xa,long xb,long ya,long yb)
{
    int x=ya/xb;
    if(ya%xb)  x++;
    if(yb-(x-1)*xa>0)
      return true;
    else
      return false;
}
int main()
{
    int t;
    long xa,xb,ya,yb;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%ld",&xa);
        scanf("%ld",&xb);
        scanf("%ld",&ya);
        scanf("%ld",&yb);
        if(ai(xa,xb,ya,yb))
          printf("YES\n");
        else 
          printf("NO\n");
    }
    return 0;
}

#include<stdio.h>
long long w[4];
int main(){
    //int Xa,Xb,Ya,Yb,T;//w 1,2,3,4
    int T;
    scanf("%d",&T);
    while(T>0){
        for(int i=1;i<=4;i++){
            scanf("%lld ",&w[i]);//输入
        }
            long long Q;//琪亚娜打的伤害Q
            Q=(w[4]/w[1]+1)*w[2];//n到没砍死琪亚娜先手所以多砍一下
            if(w[4]%w[1]==0) Q-=w[2];//芽衣n刀刚好砍死琪亚娜，所以把琪亚娜多砍的
            //一下伤害减去。
            
            if(Q>=w[3]){printf("YES\n");}
            else printf("NO\n");
        T--;
    }
}   

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
int n,x1,x2,y,y2;
int main() {
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d%d%d",&x1,&x2,&y,&y2);
        int tmp1=y2%x1,h1=y2/x1;
        int tmp2=y%x2,h2=y/x2;
        if(tmp1)
            h1++;
        if(tmp2)
            h2++;
        if(h2<=h1)
        printf("YES\n");
        else
            printf("NO\n");

    }
    return 0;
}

思路

签到题，我们只用计算琪亚娜在死之前能打出多少伤害即可，然后将其与芽衣的生命值做对比

Code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[4];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--) {
		for(int i = 0;i < 4; ++i)	{
			scanf("%lld",&a[i]);
		}
		ll keyq = (a[3]/a[0] + 1LL)*a[1];
		if(a[3] % a[0] == 0) keyq -=a[1]; //这里是特判一下如果芽衣的攻击为1的情况
		if(keyq >= a[2]) puts("YES");
		else puts("NO");
	}
	return 0;
}