107 - The Cat in the Hat

最新推荐文章于 2024-06-27 09:35:57 发布

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本文介绍了一个使用C语言实现的特殊算法，该算法通过模拟“工作猫”的行为来解决一类数学问题。具体来说，算法接受两个参数a和b，当b为1时，采用一种特定的递减方式计算与a相关的两个累计和；当b大于1时，则尝试找到能够使得b成为特定形式幂次表达式的底数N，并基于此进行进一步计算。

#include <stdio.h>
#include <stdbool.h>
#include <stdint.h>

int main ( int argc, char * argv[] ) {
	int m, n;
	bool flag = true;
	while( flag ) {
		/* a donates height,
		 * b donates number of worker cats */
		long long a, b;
		long long sum_a, sum_b;
		scanf("%lld %lld", &a, &b);
		if(a == 0 && b == 0) {
			flag = false;
		} else {

			if(b == 1) {
			
				sum_a = 0;
				sum_b = 0;

				while(a >= 1) {
					sum_a += 1;
					sum_b += a;

					a = a / 2;
				}
				sum_a = sum_a - 1;
				printf("%lld %lld\n", sum_a, sum_b);
			
			} else {
				/* N should be greater than 1 */
				int N;
				for(N = 2; N < 10000; N++) {
					long long prod = 1;
					long long last = 1;
					sum_a = last;
					int q = 0;
					while(prod < b) {
						sum_a = last;
						last = last * N + 1;
						prod = prod * N;
						q += 1;
					}

					if(prod == b) {

						int i;
						sum_b = 0;
						long long dup_a = a;
						prod = 1;
						bool tick = true;
						for(i = 0; i < (q+1); i++) {
							sum_b += dup_a * prod;
							prod = prod * N;
							if(dup_a < 1) tick = false;
							dup_a = dup_a / ( 1 + N );

						}

						if(tick) {
							printf("%lld %lld\n", sum_a, sum_b);
							N = 10000;
						}
					}	
			
				}
			}
		}
	}
	return 0;
}