（DP）codeforces - 710E Generate a String

原题链接：http://www.codeforces.com/problemset/problem/710/E

可以很容易发现，奇数肯定是+1或者通过*2逼近并且+1或-1得到。

而偶数只能在+1和翻倍得到。

所以在奇数情况下的状态转移方程为dp[i]=min(dp[i-1]+x,dp[i/2]+x+y,dp[i/2+1]+x+y)

偶数情况下的为dp[i]=min(dp[i-1]+x,dp[i/2]+y)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<set>
#include<vector>
#include<queue>
#include<map>
#include<list>
#include<bitset>
#include<string>
#include<cctype>
#include<cstdlib>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

namespace Math {
//最大公约数
ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
//欧拉素数筛
int Euler_prime(int *prime, bool *vis, int maxn) {
    memset(vis, true, sizeof(vis));
    int tot = 0;
    for (int i = 2; i < maxn; i++) {
        if (vis[i]) prime[tot++] = i;
        for (int j = 0; j < tot&&prime[j] * i < maxn; j++) {
            vis[i*prime[j]] = false;
            if (i%prime[j] == 0) break;
        }
    }
    return tot;
}
}

const int inf = 1 << 30;
const ll lnf = 1ll << 60;

//-----upstair is template------//

const int maxn=1e7;
ll dp[maxn];

void solve() {
    int n,x,y;
    scanf("%d%d%d",&n,&x,&y);
    fill(dp,dp+maxn,lnf);
    dp[0]=0;
    for(int i=1;i<=n;i++){
        if(i&1){
            dp[i]=min(dp[i-1]+x,min(dp[i/2+1]+x+y,dp[i/2]+x+y));
        }
        else{
            dp[i]=min(dp[i-1]+x,dp[i/2]+y);
        }
    }
    printf("%I64d\n",dp[n]);
}

int main() {

#ifndef ONLINE_JUDGE
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    //iostream::sync_with_stdio(false);
    solve();
    return 0;
}