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ural 1104. Don’t Ask Woman about Her Age暴力

2016-09-21 13:51 239 查看

1104. Don’t Ask Woman about Her Age

Time limit: 1.0 second
Memory limit: 64 MB

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 106 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	output
A1A	22

Problem Author: Igor Goldberg
Problem Source: Tetrahedron Team Contest May 2001

思路：从小到大暴力，注意年龄 >= 2

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <string>
using namespace std;
int N, T;
int main() {
//freopen("in.txt", "r", stdin);
string s;
cin >> s;
int cnt = 2;
for(int i = 0; i < s.size(); i++){
if(s[i] >= 'A' && s[i] <= 'Z' && s[i] - 'A' + 10 >= cnt){
cnt = s[i] - 'A' + 11;
}else if(s[i] >= '0' && s[i] <= '9'&& s[i] - '0' >= cnt){
cnt = s[i] - '0' + 1;
}
}
long long int sum = 0, now;
for(int k = cnt; k <= 36; k++){
if(s[s.size()-1] >= 'A' && s[s.size()-1] <= 'Z') sum = s[s.size()-1] - 'A' + 10;
else sum = s[s.size()-1] - '0';
sum %= (k-1);
for(int i = s.size()-2; i >= 0; i--){
if(s[i] >= 'A' && s[i] <= 'Z') now = s[i] - 'A' + 10;
else now = s[i] - '0';
sum += now*k;
sum %= (k-1);
}
if(sum == 0) {
printf("%d\n", k);
exit(0);
}
}
printf("No solution.\n");
return 0;
}

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