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

    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

    inputoutput
    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
    ", k);
                exit(0);
            }
        }
        printf("No solution.
    ");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cshg/p/5892370.html
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