Timus 1104 Don’t Ask Woman about Her Age

最新推荐文章于 2016-07-30 19:10:17 发布

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本文介绍了一个数学挑战问题：寻找一个合适的进制k，使得给定的字符串表示的数值，在该进制下可以被k-1整除。文章详细解释了解决这个问题的方法，包括如何将字符转换为数值，以及如何通过计算来确定最小的符合条件的进制。

Description

Mrs Little likes digits most of all. Every year she tries to make the best number of the year. She tries to become more and more intelligent and every year studies a new digit. And the number she makes is written in numeric system which base equals to her age. To make her life more beautiful she writes only numbers that are divisible by her age minus one. Mrs Little wants to hold her age in secret.

You are given a number consisting of digits 0, …, 9 and Latin letters A, …, Z, where A equals 10, B equals 11 etc. Your task is to find the minimal number k satisfying the following condition: the given number, written in k-based system is divisible by k−1.

Input

Input consists of one string containing no more than 10 ⁶ digits or uppercase Latin letters.

Output

Output the only number k, or "No solution." if for all 2 ≤ k ≤ 36 condition written above can't be satisfied. By the way, you should write your answer in decimal system.

Sample Input

A1A 22

题意：输入一个k base的数，能被k-1整除，求k。

思路：需要注意下列知识点要记住

（1） k base的数能除掉k-1，那么这个数各位加和也能除掉k-1

（2） k一定大于各个位的数（也就是各位的最后的max）

注意：还要考虑到特殊情况各位全为0，最小就为2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;

//将字符转化为数
int cl(char c)
{
	int r;
	if(isalpha(c)) r=c-'A'+10;   //isalpha函数判别该字符是否为字母
	else r=c-'0';
	return r;
}

int main()
{
	string a;
	int m,max_w=-1,sum=0,flag=0;
    cin>>a;
    for(int i=0;i<a.size();i++)
	{
	    m=cl(a[i]);
	    sum+=m;
	    max_w=max(max_w,m);
	}
	if(max_w==0)    //全为0的情况
	{
		cout<<"2\n"<<endl;
		return 0;
	}
	for(;max_w<36;max_w++)
	{
		if(sum%max_w==0)
		{
			flag=1;
			break;
		}
	}
	if(flag) printf("%d\n",max_w+1);
	else printf("No solution.\n");
	return 0;
}