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;
}