2019-CCPC-秦皇岛 部分题目题解

D-Decimal

题目·：Given a positive integer n, determine if 1n is an infinite decimal in decimal base. If the answer is yes, print “Yes” in a single line, or print “No” if the answer is no.
Input
The first line contains one positive integer T (1≤T≤100), denoting the number of test cases.
For each test case:
Input a single line containing a positive integer n (1≤n≤100).
Output
Output T lines each contains a string “Yes” or “No”, denoting the answer to corresponding test case.

• 签到题
• 输入一个正整数 $n$，判断 $1/n$是不是无限小数。
• 把 $n$的2、5因子除去，如果最终的结果为1则不是，反之则是。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t; cin >> t;
    while(t--)
    {
        int n; cin >> n;
        if(n % 2 == 0)
            while(n % 2 == 0) {n /= 2;}
        if(n % 5 == 0)
            while(n % 5 == 0) {n /= 5;}
        if(n == 1) cout << "No\n";
        else cout << "Yes\n";
    }
    return 0;
}

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E - MUV LUV EXTRA

题目：定义一个算式 $a*p-b*l$
给你一个无限循环小数的前几位，选择一种循环节的方案使得上式结果最大。其中 $a， b$由输入给定， $p$表示循环节已经开始出现的部分长度， $l$表示循环节的长度

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mx = 1e7 + 10;
char s2[mx], s1[mx];
int nextt[mx];
ll a, b;
void getNext(int len)
{
    //nextt数组代表当前字符之前的字符串前缀后缀最长公共元素长度（KMP）
    int i = 0, j = -1;
    nextt[0] = -1;
    while(i < len)
    {
        if(j == -1 || s2[i] == s2[j])
        {
            i++; j++;
            nextt[i] = j;
        }
        else
            j = nextt[j];
    }
}
int main()
{
    while(scanf("%d%d%s",&a,&b,s1) != EOF)
    {
        int len = 0;
        //得到小数点后的数字串
        for(int i = strlen(s1)-1; i >= 0; --i)
        {
            if(s1[i] == '.')
                break;
            s2[len++] = s1[i];
        }
        
        //得到nextt数组
        getNext(len);
        
        ll ans = a - b; //预设ans
        for(int i = 1; i <= len; ++i)
            ans = max(ans, a*i-b*(i-nextt[i])); //i-nextt[i]得到的就是最小循环节长度
        printf("%lld\n",ans);
    }
    return 0;
}

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未完···