710E - Generate a String

令dp[i]表示长度为i，最少费用。
可以发现，当且仅当i为偶数的时候，dp[i]=min{dp[i/2]+y}。当为奇数的时候，dp[i]=min{dp[i+1]+x,dp[i-1]+x}。因为这个更新直接做不好做。我们考虑倍增。因为i从0开始，一开始一定dp[1]=x。所以考虑1~2区间，2~4区间。。。。\(\frac{n}{2}\)~n，先更新偶数，在正向更新，在反向更新。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
    char ch = getchar(); ll x = 0, f = 1;
    while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e7 + 10;
int n, x, y;
ll dp[N];
int main()
{
    n = read(); x = read(); y = read();
    memset(dp, INF, sizeof(dp));
    int cur = 1;
    dp[1] = x;
    while (cur <= n)
    {
        upd(i, cur, 2 * cur)
        {
            if (i % 2 == 0)dp[i] = min(dp[i], dp[i / 2] + y);
        }
        upd(i, cur+1, 2 * cur)
        {
            dp[i] = min(dp[i], dp[i - 1] + x);
        }
        dwd(i, 2 * cur-1, cur)
        {
            dp[i] = min(dp[i], dp[i + 1] + x);
        }
        cur <<= 1;
    }
    printf("%lld\n", dp[n]);
    return 0;
}