CTU Open Contest 2019 ABF题解

链接：https://www.jisuanke.com/contest/7321?view=challenges

A  Beer Barrels

 

题意：用A和B组成一个有K位的数字，求里面出现了多少个C

思路：用组合数求解，求组合数的时候就用最简单的方法初始化就可以了。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 1010;
const int mod = 1e9 + 7;
long long C[N][N];

void init()
{
    for (int i = 0; i < N; i ++ )
        for (int j = 0; j <= i; j ++ )
            if (!j) C[i][j] = 1;
            else C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
}

int main(){
    int a, b, k, c;
    cin>>a>>b>>k>>c;
    init();
    long long sum = 0;
    if(a != c && b != c){
        cout<<0<<endl;
        return 0;
    }
    else if((a == c && b != c)||(a != c && b == c)){
        for(long long i = 0; i <= k; i ++ ){
            sum = (sum + C[k][i] * i % mod) % mod;
        }
        cout<<sum<<endl;
        return 0;
    }
    else if(a == c && b == c){
        cout<<k<<endl;
        return 0;
    }
}

B  

 

 

 题意：有一种语言用来记账，格式如题，读输入，输出该格式下的总金额

思路：把格式化到最后可得：bill = (dig_1_9 dig_0_9…，-|… 或 |… ）（bill）

代码如下：

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

int main(){
    char c[1010];
    char str[2];
    long long sum = 0;
    while(scanf("%s", c) != EOF){
        //gets(str);
        int flag = 0;
        int num = c[0] - '0';
        if(c[0] != 124){
            int i;
            for(i = 1; c[i] != ','; i ++){
                num = num * 10 + c[i] - '0';
            }
            i += 2;
            while(c[i] == 124)
                flag ++, i ++;
            sum += num * flag;
        }
        else {
            flag = strlen(c);
            sum += flag * 42;
        }
    }
    if(sum % 10 != 0)
        sum = (sum / 10 + 1) * 10;
    cout<<sum<<",-"<<endl;
    return 0;
}

F Beer Marathon

题图突然放不上来，链接：https://nanti.jisuanke.com/t/43512

 参考了题解

题意：有n个啤酒摊，要求任意两个相邻啤酒摊的距离等于给出的k值，求啤酒摊移动的总距离最少是多少

思路：二分

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1e6 + 10;
typedef long long ll;
ll a[N], n, k;

ll cal(ll w){
    ll ans = 0;
    for(ll i = 0; i < n; i ++){
        ans += abs(a[i] - w);
        w += k;
    }
    return ans;
}

int main(){
    scanf("%lld %lld", &n, &k);
    for(ll i = 0; i < n; i ++ )
        scanf("%lld", &a[i]);
    sort(a, a + n);
    ll l = -1e12, r = 1e12, mid = 0, L, R, ans1, ans2, ans = 1e18;
    for(int i = 0; i < 100; i ++){
        mid = (l + r) / 2;
        L = (mid + l) / 2;
        R = (mid + r) / 2;
        ans1 = cal(L);
        ans2 = cal(R);
        if(ans1 < ans2)
            r = R;
        else l = L;
        ans = min(ans, min(ans1, ans2));
    }
    printf("%lld
", ans);
    return 0;
}