762A k-th divisor

题目链接:

http://codeforces.com/problemset/problem/762/A

题解:

算是简单题,需要注意的就是题目中给出的数据比较大,所以,我用的是折半处理的方法,只处理平方之前的数据,在用除法把剩下的数据给处理,这里还需要注意的是i*i==n,这里需要考虑一下。

代码:

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e6+10;
#define met(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
typedef long long ll;
ll num[maxn];

int main()
{
    ll n,k;
    while(cin>>n>>k)
    {
        met(num,0);
        ll ans=-1;
        ll m=sqrt(n);
        ll cnt=0,len=0;
        for(ll i=1;i<=m;i++)
        {
            if(n%i==0)
            {
                cnt++;
                num[len++]=i;
            }
            if(cnt==k)
            {
                ans=i;
                cout<<ans<<endl;
                break;
            }
        }
        ll temp=cnt;
        if(m*m==n)
            temp--;
        if(ans==-1)
        {
            ll flag=0;
            ll i;
            for(i=len-1;i>=0;i--)
            {
                temp++;
                if(temp==k)
                {
                    flag=1;
                    break;
                }
            }
            if(flag)
                ans=n/num[i];
            cout<<ans<<endl;
        }
    }
}


#include<reg52.h> sbit ADDR0 = P1^0; sbit ADDR1 = P1^1; sbit ADDR2 = P1^2; sbit ADDR3 = P1^3; sbit ENLED = P1^4; unsigned char code LedChar[] = { 0xC0, 0xF9, 0xA4, 0xB0, 0x99, 0x92, 0x82, 0xF8, 0x80, 0x90, 0x88, 0x83, 0xC6, 0xA1, 0x86, 0x8E }; unsigned char LedBuff[6] = {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF}; unsigned char i = 0; unsigned int cnt = 0; unsigned char flagls = 0; unsigned char led_cnt = 0; // LED流水灯计数 unsigned int led_delay = 0;// LED延时计数 unsigned char mode = 0; // 工作模式:0-计时模式,1-流水灯模式 void main() { unsigned long sec = 0; unsigned long divisor; unsigned char digit; unsigned int j = 0; unsigned int k = 0; EA = 1; ENLED = 0; ADDR3 = 1; TMOD = 0x01; TH0 = 0xFC; TL0 = 0x67; ET0 = 1; TR0 = 1; while(1) { if(flagls == 1) { flagls = 0; sec++; for (j = 0; j < 6; j++) { divisor = 1; for (k = 0; k < j; k++) { divisor *= 10; } digit = (sec / divisor) % 10; if (digit != 0 || j == 6) { // Only display non-zero digits or the last digit LedBuff[j] = LedChar[digit]; } else { LedBuff[j] = 0xFF; // Keep the digit off if it's a leading zero } if(sec >= 10) { mode = 1; ADDR2 = 1; // 切换到LED显示 ADDR1 = 1; ADDR0 = 0; } } } } } void InterruotTimer0() interrupt 1 { TH0 = 0xFC; TL0 = 0x67; cnt++; if(cnt > 1000) { cnt = 0; flagls = 1; } if(mode == 0) // 计时模式 - 数码管扫描 { P2 = 0xFF; // 消隐 switch(i) { case 0: ADDR2 = 0; ADDR1 = 0; ADDR0 = 0; i++; P2 = LedBuff[0]; break; case 1: ADDR2 = 0; ADDR1 = 0; ADDR0 = 1; i++; P2 = LedBuff[1]; break; case 2: ADDR2 = 0; ADDR1 = 1; ADDR0 = 0; i++; P2 = LedBuff[2]; break; case 3: ADDR2 = 0; ADDR1 = 1; ADDR0 = 1; i++; P2 = LedBuff[3]; break; case 4: ADDR2 = 1; ADDR1 = 0; ADDR0 = 0; i++; P2 = LedBuff[4]; break; case 5: ADDR2 = 1; ADDR1 = 0; ADDR0 = 1; i = 0; P2 = LedBuff[5]; break; default: break; } } else // 流水灯模式 { led_delay++; if(led_delay >= 1000) // 1秒延时 { led_delay = 0; P2 = ~(0x01 << led_cnt); // 点亮对应LED led_cnt++; if(led_cnt >= 8) { led_cnt = 0; } } } }改进此代码,实现利用定时器T0定时10秒,并通过数码管显示秒表,到10秒后让led流水灯从左往右移动(1秒切换一个),10秒后,数码管还显示
05-18
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