Tsinsen_A1024. 瓷砖问题再讨论

4

//#include<iostream>
//#include<cstdio>
#include<stdio.h>
//蓝色部分为调试时所写 
int ans, num;
int cnt = 0, count = 0;
void dg(int x, int n1, short flag) { 
  //问题终于找出来了，对于n1的每次增加必须在确定本次所铺的瓷砖为1时， 然后在下一次进行增加，否则会导致本次可能多次自增 
	if(x == 0 && !flag) {ans++; num += n1; /*cout << "第" << ++count << "次成功------"<< "n1: " << n1 << "出栈" << endl; */return ;} 
	else if(x == 0 && flag) {n1++; ans++; num += n1;  return ;}
	//cout <<"第" << ++cnt << "次  n1: " << n1 << "  flag: " << flag << "  ans: " << ans << endl;
	if(flag) n1++;
	for(int i = 1; i <= 3; i++) 
		if(x >= i) { // cout << "i -- " << i << endl; 
			if( i == 1 && !flag) {flag = 1;/*cout << "入栈  " ;*/dg(x-i, n1, flag);} 
			else if(i == 1 && flag) ;
			else {flag = 0; /*cout << "入栈  " ;*/dg(x-i, n1, flag);} 
		} 
//	cout << " 出栈 " << endl; 一次函数的结束会减少一个栈帧，勉强写为"出栈" 
}

int main() {
	int N;
	while(scanf("%d", &N) != EOF) {
		ans = num = 0;
		dg(N, 0, 0);
		printf("%d\n%d\n", ans, num);
	}
	return 0;
}