Codeforces Round #609 (Div. 2)

题库链接

https://codeforces.com/contest/1269

A. Equation

给定一个数n,求两个合数a,b,使得a-b=n
如果n是偶数,显然找两个偶数就好了,如果n是奇数,找一个奇数,再找一个偶数

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 200000;
const int M = 1e9+7;
int n,m,k,t;
 
 
bool notprime(int x)
{
    for(int i = 2; i<= sqrt(x); i++)
    {
        if(x%i == 0) return true;
    }
    return false;
}
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin>>n;
    if(n%2)
    {
        for(int i = 4; i < maxn; i += 2)
        {
            if(notprime(i+n))
            {
                cout<<i+n<<' '<<i<<endl;
                return 0;
            }
        }
    }
    else cout<<n+10<<' '<<10<<endl;
    return 0;
}

B. Modulo Equality

2个数组,长度是n,元素上限是m,找一个最小的x,使得\((a_i + x) \bmod m = b_{p_i}\)
暴力出所有的可能的x,\(x = (b_0 - a_i + m) \bmod m\),然后判断a是否可以变成b

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 2100;
const int M = 1e9+7;
int n,m,k,t;
 
int a[maxn];
int b[maxn];
int c[maxn];
 
bool judge(int x)
{
    for(int i = 0; i < n; i++)
    {
        c[i] = (a[i]+x)%m;
    }
    sort(c,c+n);
    for(int i = 0; i < n; i++)
    {
        if(c[i] != b[i]) return false;
    }
    return true;
}
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin(n),cin(m);
    for(int i = 0; i < n; i++) 
    {
        cin(a[i]);
    }
    for(int i = 0; i < n; i++) 
    {
        cin(b[i]);
    }
    sort(b,b+n);
    set<int> st;
    for(int i = 0; i < n; i++) 
    {
        st.insert((b[0]-a[i]+m)%m);
    }
    for(auto i : st)
    {
        if(judge(i))
        {
            cout<<i<<endl;
            return 0;
        }
    }
    return 0;
}

C. Long Beautiful Integer

给定一个数a,a的长度n,循环节长度k,求一个最小的b,满足\(b_i = b_{i+k}\),并且b>a
首先判断最小的b满不满足大于a,如果不满足就b[k-1]+1,然后判断一下进位就好了,交的时候把'0'写成了'9',也是醉了

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 200100;
const int M = 1e9+7;
int n,m,k,t;
 
char s[maxn];
char ss[maxn];
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin(n),cin(k);
    cin>>s;
    for(int i = 0; i < n; i++)
    {
        ss[i] = s[i%k];
    }
    if(strcmp(ss,s) >= 0) {
        cout<<strlen(ss)<<endl;
        cout<<ss<<endl;
    }
    else
    {   
        s[k-1]++;
        int a = k-1;
        while (s[a] == '9'+1)
        {
            s[a] = '0';         //不小心写错了,自闭了
            s[a-1]++;
            a--;
        }
        for(int i = 0; i < n; i++)
        {
            ss[i] = s[i%k];
        }
        cout<<strlen(ss)<<endl;
        cout<<ss<<endl;
    }
    return 0;
}

D. Domino for Young

一个图表,第i列长度为\(a_i\),\(a_i \geq a_{i+1}\),求最多可以放下多少(2,1),或者(1,2)的多米诺骨牌
官方的题解,把这些格子染色,白色边上是黑色,黑色边上是白色,然后ans = min(白,黑)

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 300100;
const int M = 1e9+7;
int n,m,k,t;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin>>n;
    ll b = 0,w = 0;
    for(int i = 1,x; i <= n; i++) 
    {
        cin>>x;
        if(i%2)
        {
            b += (x+1)/2;
            w += (x)/2;
        }
        else 
        {
            w += (x+1)/2;
            b += (x)/2;
        }
    }
    cout<<min(b,w)<<endl;
    return 0;
}