AtCoder Grand Contest #026 B - rng_10s

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 600600 points

Problem Statement

Ringo Mart, a convenience store, sells apple juice.

On the opening day of Ringo Mart, there were AA cans of juice in stock in the morning. Snuke buys BB cans of juice here every day in the daytime. Then, the manager checks the number of cans of juice remaining in stock every night. If there are CC or less cans, DD new cans will be added to the stock by the next morning.

Determine if Snuke can buy juice indefinitely, that is, there is always BB or more cans of juice in stock when he attempts to buy them. Nobody besides Snuke buy juice at this store.

Note that each test case in this problem consists of TT queries.

Constraints

• 1≤T≤3001≤T≤300
• 1≤A,B,C,D≤10181≤A,B,C,D≤1018
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

TT
A1A1 B1B1 C1C1 D1D1
A2A2 B2B2 C2C2 D2D2
::
ATAT BTBT CTCT DTDT

In the ii-th query, A=Ai,B=Bi,C=Ci,D=DiA=Ai,B=Bi,C=Ci,D=Di.

Output

Print TT lines. The ii-th line should contain Yes if Snuke can buy apple juice indefinitely in the ii-th query; No otherwise.

---

Sample Input 1 Copy

Copy

14
9 7 5 9
9 7 6 9
14 10 7 12
14 10 8 12
14 10 9 12
14 10 7 11
14 10 8 11
14 10 9 11
9 10 5 10
10 10 5 10
11 10 5 10
16 10 5 10
1000000000000000000 17 14 999999999999999985
1000000000000000000 17 15 999999999999999985

Sample Output 1 Copy

Copy

No
Yes
No
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
Yes

In the first query, the number of cans of juice in stock changes as follows: (D represents daytime and N represents night.)

99 →D 22 →N 1111 →D 44 →N 1313 →D 66 →N 66 →D x

In the second query, the number of cans of juice in stock changes as follows:

99 →D 22 →N 1111 →D 44 →N 1313 →D 66 →N 1515 →D 88 →N 88 →D 11 →N 1010 →D 33 →N 1212 →D 55 →N 1414 →D 77 →N 77 →D 00 →N 99 →D 22 →N 1111 →D …

and so on, thus Snuke can buy juice indefinitely.

---

Sample Input 2 Copy

Copy

24
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1

Sample Output 2 Copy

Copy

No
No
No
No
No
No
Yes
Yes
No
No
No
No
Yes
Yes
Yes
No
No
No
Yes
Yes
Yes
No
No
No

首先可以直接判断的情况是a < b的情况第一天就不够买的，再就是d < b的情况，每天买b个，而当总量不多于c时才只能补充d，显然补充的量不及买的量，肯定会买完。
再就是c + 1 >= b的情况下，肯定时买不完的，因为一旦量小于等于c就会进行补充，而在量大于c的情况下总是不可能买断的。
对于一般的情况其实就是c < a - bx + dy < b 如果有解的话，就会出现不补充但是总量小于b的情况，这是肯定可以买完的，否则就是永远买不完的。
对于这个不等式，移项得到a - b < bx - dy < a - c，如果bx - dy在这个区间内有解，那么区间内存在一个值能被gcd(b,d)整除，但是这个区间是开区间(a - b,a - c),
具体判断存在的方法是，如果a - b < t < a - c,t % gcd(b,d) == 0，那么就是可以买完的("No")，那么a - c - 1 >= t,显然对于整型来说t / gcd(b,d) - (a - b) / gcd(b,d) > 0,
也就是(a - c - 1) / gcd(b,d) - (a - b) / gcd(b,d) > 0,否则就是买不完的("Yes").
java代码：

import java.util.*;

public class Main {
    static long gcd(long a,long b) {
        if(b == 0)return a;
        return gcd(b,a % b);
    }
    static boolean check(long a,long b,long c,long d) {
        if(a < b || d < b)return false;
        if(c + 1 >= b)return true;
        ///如果某天之后  剩下的多于c(不需要增加）  但是却小于b 那么次日肯定不够
        ///实际上就是c < a - bx + dy < b有解的话就是No
        ///移项的a - b < bx - dy < a - c
        ///根据欧几里德扩展定理 就是说(a-b,a-c)之间是否有gcd(b,d)的倍数 即是否有解
        long g = gcd(b,d);
        return (a - c - 1) / g - (a - b) / g <= 0;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for(int i = 0;i < t;i ++) {
            long a = in.nextLong();
            long b = in.nextLong();
            long c = in.nextLong();
            long d = in.nextLong();
            if(check(a,b,c,d))System.out.println("Yes");
            else System.out.println("No");
        }
    }

}

c代码：

#include <stdio.h>
long long gcd(long long a,long long b){
    return b ? gcd(b,a % b) : a;
}
int check(long long a,long long b,long long c,long long d){
    if(a < b || d < b)return ;
    if(c +  >= b)return ;
    long long g = gcd(b,d);
    return (a - c - ) / g - (a - b) / g <= ;
}
int main() {
    int t;
    long long a,b,c,d;
    scanf("%d",&t);
    while(t --){
        scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
        if(check(a,b,c,d)){
            printf("Yes\n");
        }
        else{
            printf("No\n");
        }
    }
}