Codeforces 251C Number Transformation

Number Transformation

我们能发现这个东西是以2 - k的lcm作为一个循环节， 然后bfs就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 5e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

LL a, b;
int k, lcm = ;
int dp[N];

void getDp(int x) {
    memset(dp, -, sizeof(dp));
    dp[x] = ;
    queue<int> que;
    que.push(x);
    while(!que.empty()) {
        int u = que.front(); que.pop();
        if(dp[u - ] == -) {
            dp[u - ] = dp[u] + ;
            que.push(u - );
        }
        for(int i = ; i <= k; i++) {
            int v = u - (u % i);
            if(~dp[v]) continue;
            dp[v] = dp[u] + ;
            que.push(v);
        }
    }
}

int main() {
    scanf("%lld%lld%d", &a, &b, &k);
    swap(a, b);
    for(int i = ; i <= k; i++)
        lcm = lcm / __gcd(lcm, i) * i;
    LL ans = ;
    LL R = b - (b % lcm);
    LL L = a + (lcm - a % lcm) % lcm;
    if(L <= R) {
        LL cnt = (R - L) / lcm;
        getDp(lcm - );
        ans = dp[] * cnt + cnt;
        if(a % lcm) ans += dp[a % lcm] + ;
        getDp(b % lcm);
        ans += dp[];
    } else {
        getDp(b % lcm);
        ans = dp[a % lcm];
    }
    printf("%lld\n", ans);
    return ;
}

/*
*/