4 solutions
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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--; } }
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0
#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; }
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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; }
- 1
Information
- ID
- 265
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 8
- Tags
- # Submissions
- 898
- Accepted
- 112
- Uploaded By