Problem E 不妙(From 2019计科第二次实训)

10175101159 edited 2 年前

本弱鸡还不会数位DP,于是使用了一些数学技巧进行正面实现(懒得写思路了,大概是排列组合):

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

string a,b;

ll mpow(ll a,ll b)
{
    ll ret = 1;
    while(b--){
        ret *= a;
    }
    return ret;
}

ll mix(string s)
{
    ll temp = 0;
    for(int i = s.length() - 2;i >= 0;--i){
        temp += s[i] - '0';
    }
    ll ret = temp % 9;
    return ret ? 9 - ret : 0;
}

ll cal(string s)
{
    ll len = s.length();
    ll ret = 0;
    for(ll i = 0;i < len - 1;++i){
        if(s[i] == '9'){
            for(ll j = i;j < len;++j){
                s[j] = '8';
            }
        }
        ret += (s[i] - '0') * mpow(9LL,len - i - 2) * 8LL;
    }
    ll left = min(s[len - 1],'8') - '0';
    ret += left + 1;
    if(mix(s) <= left)--ret;
    return ret;
}

int main()
{
    while(cin>>a>>b){
        cout<<cal(b) - cal(a) + 1<<endl;
    }
    return 0;
}

这题还可以放宽条件,不要求输入的$a,b$都是妙数,代码如下:

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

ll a,b;

string change(ll x)
{
    if(x == 0)return "0";
    string ret = "";
    while(x){
        ret = (char)(x % 10 + '0') + ret;
        x /= 10;
    }
    return ret;
}

ll mpow(ll a,ll b)
{
    ll ret = 1;
    while(b--){
        ret *= a;
    }
    return ret;
}

ll mix(string s)
{
    ll temp = 0;
    for(int i = s.length() - 2;i >= 0;--i){
        temp += s[i] - '0';
    }
    ll ret = temp % 9;
    return ret ? 9 - ret : 0;
}

ll cal(string s)
{
    ll len = s.length();
    ll ret = 0;
    for(ll i = 0;i < len - 1;++i){
        if(s[i] == '9'){
            for(ll j = i;j < len;++j){
                s[j] = '8';
            }
        }
        ret += (s[i] - '0') * mpow(9LL,len - i - 2) * 8LL;
    }
    ll left = min(s[len - 1],'8') - '0';
    ret += left + 1;
    if(mix(s) <= left)--ret;
    return ret;
}

int main()
{
    while(cin>>a>>b){
        cout<<cal(change(b)) - cal(change(a - 1))<<endl;
    }
    return 0;
}

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