uva 10127 - Ones(数论)

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uva 10127 - Ones(数论)

题目链接:uva 10127 - Ones 题目大意:给出n,问说者少要多少为1才干够整除n. 解题思路:等于是高精度取模,直到余数为0为止. #include <cstdio> #include <cstring> int main () { int n; while (scanf("%d", &n) == 1) { int ans = 1, c = 1; while (c) { c = (c * 10 + 1) % n; ans++; } print…

【数论，水题】UVa 10127 - Ones

题目链接题意:给你一个数n,问最少有多少个1构成的“1”串(1,11,...)能整除n; 比如:111能被3整除: 111111能被7整除:... 作为水货觉得只要自己能1A的都是水题=. = #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> using namespace std; ; int main() { int n, p[maxn]; while(~scan…

uva 1434 - YAPTCHA(数论)

题目链接:uva 1434 - YAPTCHA 题目大意:给定n和k,求题目中给定的式子S(n). 解题思路:威尔逊定理,x为素数时有,((x−1)!+1)%x==0,所以对于本题.假设3*k+7为素数的话,[(3k+6)!+1(3k+7−[(3k+6)!3k+7]]=1 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std…

UVA 11645 - Bits(数论+计数问题)

题目链接:11645 - Bits 题意:给定一个数字n.要求0-n的二进制形式下,连续11的个数. 思路:和 UVA 11038 这题相似,枚举中间,然后处理两边的情况. 只是本题最大的答案会超过longlong,要用高精度,只是借鉴http://www.cnblogs.com/TO-Asia/p/3214706.html这个人的方法,直接用两个数字来保存一个数字.这样能保存到2个longlong的长度,就足够存放这题的答案了. 代码: #include <stdio.h> #include…

UVA 10127题目的解答

#include <iostream>#include <cstdio>#include <cmath> int main(){ int num; while (scanf_s("%d",&num) != EOF) { int cnt = 2; long long testNum = 11; while (true) { if (testNum%num == 0) { break; } testNum = testNum * 10 + 1;…

UVA 10127题目描述

Given any integer 0 ≤ n ≤ 10000 not divisibleby 2 or 5, some multiple of n is a number whichin decimal notation is a sequence of 1’s. Howmany digits are in the smallest such a multipleof n?InputA ﬁle of integers at one integer per line.OutputEach out…