Codeforces Round #609 (Div. 2) A到C题

签到，乘以两个相邻的合数

#include<bits/stdc++.h>

using namespace std;

int main(int argc, char const *argv[])
{
    long long n;
    cin>>n;
    cout<<9*n<<" "<<8*n<<endl;
    return 0;
}

链接

这里的加数取模相当于移位，把两个序列排序后，遍历移位对应情况就可以。

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;

    ll A[20010];
    ll B[40010];
    int main () {
        ll n,m;

        cin>>n>>m;
        for(int i = 0; i < n; i++) {
            cin>>A[i];
        }
        for(int i = 0; i < n; i++) {
            cin>>B[i];
        }
        sort(A, A+n);
        sort(B, B+n);
        for(int i = 0; i < n; ++i){
            B[i + n] = B[i]; 
        }
        ll ans = (ll)1<<63-1;
        for(ll i = 0; i < n; ++i){
            ll diff = (B[i]-A[0]+m)%m;
            bool flag = 1;
            for(int j = 0; j < n; ++j){
                if((A[j]+diff)%m != B[i+j]) {
                    flag = 0;
                    break;
                }
            }
            if(flag){
                ans = min(ans, diff);
            }
        }
        cout << ans << endl;
    }

链接

取前k个，后面的数，要么是前k个的重复出现，要么是从第k个往前，是如果是9则变零，找到第一个不是9的加一。重复输出就可以。

#include<bits/stdc++.h>

using namespace std;

char a[200010];
char b[200010];
int main()
{
    int m,k;
    cin>>m>>k;
    //scanf("%s",a);
    int flag = 1;
    for (int i = 0; i < m; ++i)
    {
        cin>>a[i];
    }
    //memset(b, 0, sizeof(b));
    for (int i = 0; i < m; ++i)
    {
        if (a[i%k]<a[i])
        {
            flag = 0;
            break;
        }
        if(a[i%k]>a[i])
        {
            flag = 1;
            break;
        }
    }
    //用来记录重叠的部分
    for (int i = 0; i < k; ++i)
    {
        b[i] = a[i];
    }
    if (flag)
    {
        cout<<m<<endl;
        for (int i = 0; i < m; ++i)
        {
            cout<<b[i%k];
        }
        cout<<endl;
    } 
    else
    {
        //先增加， 从尾部开始；
        for (int i = k-1; i >=0 ; --i)
        {
            if (b[i]!='9')
            {
                b[i] += 1;
                //flag = 1;
                break;
            }
            else
            {
                b[i]='0';
            }
        }
        cout<<m<<endl;
        for (int i = 0; i < m; ++i)
        {
            cout<<b[i%k];
        }
        cout<<endl;
        
            
    }
    return 0;
}