正睿20NOIp前冲刺day13

估分：100+40+35+30=205

实际：100+20+15+30=175

T1:

　　数位DP，打表就可以发现规律

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long LL;

inline LL read()
{
    LL x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f = ch == '-', ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
}

const int popc[300] = { 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8 };

int popcount(LL x) { return popc[x & 0xff] + popc[(x >> 8) & 0xff] + popc[(x >> 16) & 0xff] + popc[(x >> 24) & 0xff] + popc[(x >> 32) & 0xff] + popc[(x >> 40) & 0xff] + popc[(x >> 48) & 0xff] + popc[(x >> 56) & 0xff]; }

struct Exactly
{
    LL f[110];

    void work(LL n)
    {
        f[0] = 1;
        for (int i = 1; i <= 65; i++) f[i] = 2 * f[i - 1] + 1;

        LL ans = 0;
        for (int i = 0; n; n >>= 1, i++) if (n & 1) ans += f[i];
        printf("%lld", ans);
    }
}E;

int main()
{
    LL n = read();
    if (n <= 1000000)
    {
        LL ans = 0;
        for (LL i = 0; i < n; i++) ans += popcount(i ^ (i + 1));
        printf("%lld", ans);
    }
    else E.work(n), exit(0);
}

T2:

　　脑抽把n=1的情况改了（什么时候改的都不知道），丢了20pts

　　先看n=2,x≤1000的情况，f(i,j)表示第一种颜色有i个球，第二种颜色有j个球，用一个筹码能换来多少个筹码，首先f(0,1)=f(1,0)=2。然后一个f(i,j)，可以拿一个球变成f(i-1,j)或f(i,j-1)，但要最优情况，所以只有无论变成哪种情况，最后得的筹码都不变才最优，否则幕后黑手一定会把状态变成最后得的筹码最少的状态。假设a=f(i-1,j),b=f(i,j-1)，押了x个筹码在第一种颜色的球上，剩下的(1-x)就押在第二种颜色的球上，我们就要求ax=b(1-x)，则x=b/(a+b)，那么f(i,j)就等于2*b/(a+b)*a。

　　之后看优化这个DP，设$g(i,j)=\frac{f(i,j)}{2^{i+j}}$，那么g的转移方程就是$g(i,j)=\frac{g(i-1,j)g(i,j-1)}{g(i-1,j)+g(i,j-1)}$，两边取倒后就变成$\frac{1}{g(i,j)}=\frac{g(i-1,j)g(i,j-1)}{g(i-1,j)+g(i,j-1)}=\frac{1}{g(i-1,j)}+\frac{1}{g(i,j-1)}$，可以发$\frac{1}{f(i,j)}$其实就是在平面上走，只能往上或往右走，从(0,0)走到(i,j)的方案数，也就是$\binom{i+j}{i}$，答案就是$\frac{2^{i+j}}{\binom{i+j}{i}}$

　　之后再考虑n>2的时候怎么做，也可以写出和n=2一样的转移方程，只不过是n维的，设$f(x_{1},x_{2},x_{3},x_{4},x_{5},...,x_{n})$表示从(0,0,0,0,0,...,0)走到$x_{1},x_{2},x_{3},x_{4},x_{5},...,x_{n}$的方案数，所以$f(x_{1},x_{2},x_{3},x_{4},x_{5},...,x_{n})=\frac{\left (\sum_{i=1}^{n}x_{i} \right )!}{\prod_{i=1}^{n}x_{i}!}$，之后就能算出答案了。

　　搞不懂他们写的贪心为什么能过

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

inline int read()
{
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f = ch == '-', ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
}

const int N = 1000010;
const int mod = 998244353;

int n;
int a[N], fac[N], inv[N];

inline int qpow(int a, int k, int mod)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        k >>= 1;
    }
    return res;
}

void init(int s)
{
    fac[0] = 1;
    for (int i = 1; i <= s; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
    inv[s] = qpow(fac[s], mod - 2, mod);
    for (int i = s - 1; ~i; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}

int main()
{
    n = read();
    int sum = 0;
    for (int i = 1; i <= n; i++) a[i] = read(), sum += a[i];

    init(sum);

    int ans = 1ll * qpow(n, sum, mod) * inv[sum] % mod;

    for (int i = 1; i <= n; i++) ans = 1ll * ans * fac[a[i]] % mod;
    printf("%d", ans);
}

T3:

　　也不知道暴力哪错了，换种写法就对了，暴力去了20pts

T4:

　　30pts暴力写完就跑

总结：

　　以后还是留个十几分钟多检查下代码吧。二三题都是推出正解的公式但不会算，之后要多补数学。还有为什么二题贼简单的贪心能过啊，感性证明吗