BUPT OJ210 Rist-Number_北邮计算机学院大二下数据结构课程设计旅游-优快云博客

本文介绍了一种名为 Rist-Number 的整数集合及其判定方法。该集合包含满足特定条件的整数，例如能被 1、3、7、15 或 31 整除的数。通过预处理的方式建立了一个标记数组来快速判断一个给定的整数 n 是否属于此集合。

题目描述

Rist-Number is a kind of integers that should satisfy some restrictions. Define S as the set of all Rist-Numbers, then we have

1∈S
3×S∈S
7×S∈S
15×S∈S
31×S∈S

Obviously, S is an infinite set. Given n, judge whether n∈S is true.

输入格式

The input contains multiple test cases.

The first line of input contains an integer T(T≤100), which denotes the number of test cases.

The following T lines describe all the queries, each with an integer n(1≤n≤10000).

输出格式

For each test case, output True if the statement is true, otherwise output False.

输入样例

2
1
2

输出样例

True
False

不知为何用除法判断是否只含有3,7,15,31的素因数会WA, 可能是OJ数据有问题吧.

/*
USER_ID: test#birdstorm
PROBLEM: 210
SUBMISSION_TIME: 2014-04-01 12:35:16
*/
#include <stdio.h>
#include <stdlib.h>
#define For(i,m,n) for(i=(m);i<(n);i++)
#define MAXN 100005
  
int a[MAXN];
  
main()
{
    int t, n;
    scanf("%d",&t);
    a[1]=1;
    For(n,1,3000) if(a[n]) a[n*3]=a[n*7]=a[n*15]=a[n*31]=1;
    while(t--){
        scanf("%d",&n);
        if(a[n]==1) puts("True");
        else puts("False");
    }
    return 0;
}