题解 | Moon Mooing-牛客假日团队赛14F题

题目描述

A full moon casts some sort of spell on the cows and, like their cousins the wolves and coyotes, they bay at the moon -- mooing instead of howling, of course. 

Each 'moo' lasts a certain amount of time. A short 'moo' might last time 1; a longer one might last time 24 or even 1,000,000,000 or longer (cows can really moo when they want to). No 'moo' will last more than or equal to 2^63. 

It should come as no surprise that the cows have a pattern to their moos. Bessie will choose an integer c (1 <= c <= 100) that is the initial length of a moo. After Bessie moos for length c, the cows calculate times for subsequent moos. They apply two formulae to each moo time to yield even more moo times. The two formulae are: 

 f1(c)=a1*c/d1+b1 (integer divide, of course) and 

 f2(c)=a2*c/d2+b2. 

They then successively use the two new times created by evaluating f1(c) and f2(c) to create even more mooing times. They keep a sorted list of all the possible mooing times (discarding duplicates). 

They are allowed to moo a total of N times (1 <= N <= 4,000,000). Please determine the length of the longest moo before they must quit. 

The constants in the formulae have these constraints: 1 <= d1 < a1; d1 < a1 <= 20; 0 <= b1 <= 20; 1 <= d2 < a2; d2 < a2 <= 20; 0 <= b2 <= 20.

Consider an example where c=3 and N=10. The constants are: 

 a1=4 b1=3 d1=3 

 a2=17 b2=8 d2=2 

The first mooing time is 3, given by the value of c. The total list of mooing times is: 

 1. c=3 -> 3                             6. f2(3)=17*3/2+8  -> 33 

 2. f1(3)=4*3/3+3 -> 7             7. f1(28)=4*28/3+3 -> 40 

 3. f1(7)=4*7/3+3 -> 12           8. f1(33)=4*33/3+3 -> 47 

 4. f1(12)=4*12/3+3 -> 19      9. f1(40)=4*40/3+3 -> 56 

 5. f1(19)=4*19/3+3 -> 28    10. f1(47)=4*47/3+3 -> 65 

The tenth time is 65, which would be the proper answer for this set of inputs.

65

#include <cstdio>
#include <iostream>
typedef unsigned long long ull; //注意本题要用unsigned long long。
const int N = 4000000; 
ull a[N+5]; //这里我们将a数组看成一个队列
inline ull mn(ull x, ull y) {
    return x < y ? x : y;
}
int main() {
    int c, n, a1, b1, d1, a2, b2, d2, i = 1, f1 = 1, f2 = 1;
    scanf("%d%d%d%d%d%d%d%d", &c, &n, &a1, &b1, &d1, &a2, &b2, &d2);
    a[i++] = c; //初始元素c入队
    while(i <= N) {
        ull x = mn(a1*a[f1]/d1+b1, a2*a[f2]/d2+b2); //取F1()和F2()中的较小值
        a[i++] = x; //将该较小值入队
        if(x == a1*a[f1]/d1+b1) f1++; //如果较小值来自F1()，则将F1()的指针f1+1。
        if(x == a2*a[f2]/d2+b2) f2++; //如果较小值来自F2()，则将F2()的指针f2+1
    } //重点理解while循环的内容。因为算出的F1()或F2()的数据单调递增，所以可以用这样的方式生成整个数列
    std::cout << a[n]; //最后输出即可。
    return 0;
}
//祝各位早日AC此题！