Uva--350 (数学)

2014-06-11 13:59:56

题意＆思路：简单题不多说。

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

int main(){
    int s[10005],p[10005],Z,I,M,L,Case = 0;
    while(scanf("%d %d %d %d",&Z,&I,&M,&L) == 4){
        if(Z == 0 && I == 0 && M == 0 && L == 0)
            break;
        memset(s,0,sizeof(s));
        memset(p,0,sizeof(p));
        s[L] = 1;
        p[L] = 1;
        for(int cnt = 2; ; ++cnt){
            L = (Z * L + I) % M;
            if(s[L]){
                printf("Case %d: %d\n",++Case,cnt - p[L]);
                break;
            }
            else{
                s[L] = 1;
                p[L] = cnt;
            }
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/naturepengchen/articles/3781821.html