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Uva 202 Repeating Decimals

时间:2015-02-28 21:26:03      阅读:146      评论:0      收藏:0      [点我收藏+]

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关键是判断循环的长度,若某一步的余数已出现过,则循环确定
#include <cstdio> #include <cstring> int s[10000], y[10000],sign[10000]; int main() { int a, b, sz, yu, k , cnt; while(scanf("%d%d",&a,&b)==2) { memset(sign,0,sizeof(sign)); getchar(); k = 0, cnt = 0; sz = a / b; yu = a % b; y[0] = yu; while(sign[yu] != 1) { s[cnt++] = yu * 10 / b; sign[yu] = 1; yu = yu * 10 % b; y[cnt] = yu; } printf("%d/%d = %d.",a,b,sz); for(int i = 0; i < cnt && i < 50; i++) { if(y[i] == yu) { k = i; printf("("); } printf("%d",s[i]); } if((cnt - k) > 50) printf("..."); printf(")\n %d = number of digits in repeating cycle\n\n",cnt - k); } return 0; }

 

Uva 202 Repeating Decimals

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原文地址:http://www.cnblogs.com/ekinzhang/p/4306188.html

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