codeforce #582 div3补题A - Chips MovingB - Bad PricesC - Book ReadingD1 - Equalizing by Division (easy version)D2 - Equalizing by Division (hard version)

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我是靠谱客的博主俭朴小土豆，最近开发中收集的这篇文章主要介绍codeforce #582 div3补题A - Chips MovingB - Bad PricesC - Book ReadingD1 - Equalizing by Division (easy version)D2 - Equalizing by Division (hard version)，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

A - Chips Moving

数奇数偶数的个数，取最小

//
//  main.cpp
//  cf_#582_div3_补题
//
//  Created by 陈冉飞 on 2019/9/6.
//  Copyright © 2019 陈冉飞. All rights reserved.
//
 
#include <iostream>
using namespace std;
int a[105];
 
int main(int argc, const char * argv[]) {
    int n,tem,cnt1 = 0,cnt2 = 0;
    scanf("%d",&n);
    for (int i = 0; i < n; i++) {
        scanf("%d",&tem);
        if (tem%2 == 0) cnt1++;
        else cnt2++;
    }
    cout<<min(cnt1,cnt2)<<endl;
    return 0;
}

B - Bad Prices

反向遍历，只要遍历到的数大于后面的数，就ans++，然后每次刷新从最后到当前的最小的数。

//
//  main.cpp
//  B
//
//  Created by 陈冉飞 on 2019/9/6.
//  Copyright © 2019 陈冉飞. All rights reserved.
//
 
#include <iostream>
using namespace std;
int T,n,a[150050];
 
int main(int argc, const char * argv[]) {
    for (scanf("%d",&T); T; T--) {
        scanf("%d",&n);
        int ans = 0;
        for (int i = 0; i < n; i++) scanf("%d",&a[i]);
        int stdn = a[n-1];
        for (int i = n-1; i >= 0; i--) {
            if (a[i] >stdn) ans++;
            stdn = min(stdn,a[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}

C - Book Reading

直接遍历数据大会超时，然后每次变化相同的数尾数肯定符合规律，必然找到循环节，标记循环节的长度大小最后再遍历比循环节小的那部分。

//
//  main.cpp
//  C
//
//  Created by 陈冉飞 on 2019/9/6.
//  Copyright © 2019 陈冉飞. All rights reserved.
//
 
#include <iostream>
using namespace std;
typedef long long ll;
int T;
ll n,m;
 
int main(int argc, const char * argv[]) {
    for (scanf("%d",&T); T; T--) {
        scanf("%lld%lld",&n,&m);
        ll ans = 0;
        if (m > n) ans = 0;
        else if (m == n) ans = m%10;
        else {
            //找寻环节
            ll tem = m;
            while (tem <= n) {
                ans += tem%10;
                tem += m;
                if (tem%10 == m%10) {
                    tem -= m;
                    break;
                }
            }
            ans *= (n/tem);
            tem *= (n/tem);
            while (tem <= n) {
                ans += tem%10;
                tem += m;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

D1 - Equalizing by Division (easy version)

用二维向量储存数据，数据表示的是所有的数中变化到0，1，2……的个数，然后分别是多少步，所以最后遍历的时候只需要看一下能变化到一个数的个数大于等于k的，然后筛选出变化到一个数的最小三个步数，得到变到这个数的步数，然后最后得到了所有可以得到一个数的步数，选最小的即可。

//
//  main.cpp
//  D
//
//  Created by 陈冉飞 on 2019/9/6.
//  Copyright © 2019 陈冉飞. All rights reserved.
//
 
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<vector<int>> vb(200050);   //vb用来储存所有的未除到0的前面的所有的次数
 
int main(int argc, const char * argv[]) {
    int n,k,tem;
    scanf("%d%d",&n,&k);
    for (int i = 0; i < n; i++) {
        scanf("%d",&tem);
        int cnt = 0;
        vb[tem].push_back(0);
        while (tem > 0) {
            cnt++;
            tem /= 2;
            vb[tem].push_back(cnt);
        }
    }
    int ans = 0x3f3f3f;
    for (int i = 0; i < 200050; i++) {
        if (vb[i].size()>=k) {
            int tem = 0;
            sort(vb[i].begin(), vb[i].end());
            for (int j = 0; j < k; j++) {
                tem += vb[i][j];
            }
            ans = min(ans,tem);
        }
    }
    cout<<ans<<endl;
    return 0;
}

D2 - Equalizing by Division (hard version)

只是数据量变大

//
//  main.cpp
//  D
//
//  Created by 陈冉飞 on 2019/9/6.
//  Copyright © 2019 陈冉飞. All rights reserved.
//
 
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<vector<int>> vb(200050);   //vb用来储存所有的未除到0的前面的所有的次数
 
int main(int argc, const char * argv[]) {
    int n,k,tem;
    scanf("%d%d",&n,&k);
    for (int i = 0; i < n; i++) {
        scanf("%d",&tem);
        int cnt = 0;
        vb[tem].push_back(0);
        while (tem > 0) {
            cnt++;
            tem /= 2;
            vb[tem].push_back(cnt);
        }
    }
    int ans = 0x3f3f3f;
    for (int i = 0; i < 200050; i++) {
        if (vb[i].size()>=k) {
            int tem = 0;
            sort(vb[i].begin(), vb[i].end());
            for (int j = 0; j < k; j++) {
                tem += vb[i][j];
            }
            ans = min(ans,tem);
        }
    }
    cout<<ans<<endl;
    return 0;
}