牛客OI赛制测试赛3A数字权重

版权声明：转就转吧~~记得声明噢~~ https://blog.csdn.net/Soul_97/article/details/82683537

传送

看例子可以推出来 当n=2时，答案于k的关系为  

k > 0   ans = 9 - k

k < 0  ans = 10 + k

k = 0 ans = 9

n每增多一位  答案乘10  

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 105;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
ll quickpow(ll a,ll b)
{
    ll ans = 1;
    while(b){
        if(b&1)
            ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
int main()
{
    ll n, k, ans;
    cin >> n >> k;

    if(k > 0)
        ans = 9 - k;
    else if(k < 0)
        ans = 10 + k;
    else
        ans = 9;

    ll t = quickpow(10, n - 2) % mod;
    ll sum = 0;

    for(int i = 1;i <= ans;i ++)
        sum = (sum + t) % mod;
    cout << sum << '\n';
    return 0;
}