HDU 1099 Lottery (数学问题模拟)

最新推荐文章于 2019-09-25 14:37:03 发布

Sun_AC

最新推荐文章于 2019-09-25 14:37:03 发布

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分类专栏：模拟 HDU 文章标签： hdu 数学模拟

本文链接：https://blog.youkuaiyun.com/YJX_xx/article/details/37692169

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该博客探讨了HDU 1099 Lottery问题，通过数学分析得出公式n/n+n/(n-1)+n/(n-2)+…+n/2+n/1，并指出在较小范围内可采用暴力打表的方法求解。

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Lottery

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2420 Accepted Submission(s): 1097

Problem Description

Eddy's company publishes a kind of lottery.This set of lottery which are numbered 1 to n, and a set of one of each is required for a prize .With one number per lottery, how many lottery on average are required to make a complete set of n coupons?

Input

Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=22, giving the size of the set of coupons.

Output

For each input line, output the average number of lottery required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of ouput.

Sample Input

2
5
17

Sample Output

有题意可推出n/n+n/(n-1)+n/(n-2)+……n/2+n/1

AC代码：

#include <stdio.h>
#define ll __int64
//author:XXYY
ll gcd(ll a,ll b)
{
	if(b==0)
		return a;
	return gcd(b,a%b);
}
int main(){
	ll n,i,m,z,len1,len2,x,n1,n2,y;
	while(~scanf("%I64d",&n)){
		m=0;
		z=1;
		for(i=n;i>0;i--){//通分
			m=n*z+m*i;
			z*=i;
			x=gcd(m,z);
			m/=x;
			z/=x;
		}
		if(m%z){
			y=m/z;
			m=m%z;
			n1=y;
			len1=len2=0;
			while(n1){
				len1++;
				n1/=10;
			}
			n2=z;
			while(n2){
				len2++;
				n2/=10;
			}
			for(i=0;i<=len1;i++)
				printf(" ");
			printf("%I64d\n",m);
			printf("%I64d ",y);
			for(i=0;i<len2;i++)
				printf("-");
			printf("\n");
			for(i=0;i<=len1;i++)
				printf(" ");
			printf("%I64d\n",z);
		}
		else
			printf("%I64d\n",m/z);
	}
	return 0;
}

看到这么小的范围，暴力打表也是可以的：

#include<stdio.h>
int main(){
	int n;
	while(~scanf("%d",&n)){
		switch(n){
			case 1:
				printf("1\n");
				break;
			case 2:
				printf("3\n");
				break;
			case 3:
				printf("  1\n5 -\n  2\n");
				break;
			case 4:
				printf("  1\n8 -\n  3\n");
				break;
			case 5:
				printf("   5\n11 --\n   12\n");
				break;
			case 6:
				printf("   7\n14 --\n   10\n");
				break;
			case 7:
				printf("   3\n18 --\n   20\n");
				break;
			case 8:
				printf("   26\n21 --\n   35\n");
				break;
			case 9:
				printf("   129\n25 ---\n   280\n");
				break;
			case 10:
				printf("   73\n29 ---\n   252\n");
				break;
			case 11:
				printf("   551\n33 ----\n   2520\n");
				break;
			case 12:
				printf("   551\n37 ----\n   2310\n");
				break;
			case 13:
				printf("   9473\n41 -----\n   27720\n");
				break;
			case 14:
				printf("   13433\n45 -----\n   25740\n");
				break;
			case 15:
				printf("   18581\n49 -----\n   24024\n");
				break;
			case 16:
				printf("   4129\n54 -----\n   45045\n");
				break;
			case 17:
				printf("   340463\n58 ------\n   720720\n");
				break;
			case 18:
				printf("   620743\n62 ------\n   680680\n");
				break;
			case 19:
				printf("   1662439\n67 -------\n   4084080\n");
				break;
			case 20:
				printf("   3704479\n71 -------\n   3879876\n");
				break;
			case 21:
				printf("   408335\n76 ------\n   739024\n");
				break;
			case 22:
				printf("   46533\n81 ------\n   235144\n");
				break;
		}
	}
	return 0;
}