逆元用于用于对除法运算取模的运算

$\frac{a}{b} mod $ $ m = a \cdot inv(b) mod $ $ m$

$inv(x)$ 表示 $x$ 的逆元

$inv(x) \cdot x \equiv 1 (mod $ $m)$

求解逆元

方法一：拓展欧几里得求逆元（求单个数的逆元）

已知 $\\a\cdot x + b\cdot y = 1$

同时取模 $b$ 得 $ax \equiv 1(mod $ $b)$

可见 $x$ 是 $a$ 的逆元

所以 $inv(x) = ex\_gcd(a, b, x, y)$ 中的 $x$

int ex_gcd(int a, int b, int &x, int &y){if (b == 0){x = 1;y = 0;return a;}int tmp = ex_gcd(b, a % b, y, x);y -= a/b*x;return tmp;
}int inv(int a, int mod){ //a对于mod的逆元int x, y;ex_gcd(a, mod, x, y);return x;
}

方法二：费马小定理（求单个数的逆元）

$\\x^{p-1} \equiv 1 (mod $ $ p) \\x\cdot x ^{p-2} \equiv 1(mod $ $ p)$

可见 $x^{p-2}$ 是 $x$ 的逆元

int fp(int x, int p){if (p == 1) return 1 % mod;int tmp = fp(x, p / 2);if (p & 1) return tmp * tmp % * x % mod;return tmp * tmp % mod;
}int inv(int a, int mod){return fp(a, mod - 2);
}

方法三：线性求逆元（求一段区间内的所有逆元）

$inv(x) = (m -\frac{m}{i})inv(m\%i) \%m$

假设 $t = \frac{m}{i} , k = m \% i$ ;
$\\t * i + k \equiv 0 ( mod m ) \\- t * i \equiv k ( mod m ) \\- t * 1 / k \equiv 1 / i ( mod m )\\-t*inv(k) \equiv inv(i) (mod m)$
再替换 t，k
$inv [ i ] = ( m - m / i ) * inv [ m \% i ] \% m$

inv[1] = 1;
for(int i = 2; i <= n ; i++)  inv[i] = ( m - m / i ) * inv[ m % i ] % m;

在求解组合数的时候，分子通常会出现某个数的阶乘，下面讲解阶乘逆元

方法一：正推（同时用到 $inv(x) inv(x!)$ ）

将 $inv(x!)$ 记为 $inv\_fac(x)$

$inv\_fac(i) = inv\_fac(i-1)* inv(i)$

inv[1]=inv_fac[i]=1;
for(int i=2;i<N;i++){  inv[i]=(mod-mod/i)*inv[mod%i]%mod;  //求inv[i]inv_fac[i]=inv[i]*inv_fac[i-1]%mod;
}

方法二：逆推（只用到 $inv(x!)$ ）

先用费马小定理或者拓展欧几里得先求出 $inv(n!)$

$\\\because inv(x!) = inv((x-1)!)*inv(x) \\\therefore inv(x!)*x= inv((x-1)!)$

    inv_fac[N] = fp(fac[N], mod - 2ll);for (int i = 999; i >= 1; i--){inv_fac[i] = inv_fac[i + 1] * (i + 1) % mod;}

例题HDU5651

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2981 Accepted Submission(s): 993

Problem Description

As we all known, xiaoxin is a brilliant coder. He knew **palindromic** strings when he was only a six grade student at elementry school.

This summer he was working at Tencent as an intern. One day his leader came to ask xiaoxin for help. His leader gave him a string and he wanted xiaoxin to generate palindromic strings for him. Once xiaoxin generates a different palindromic string, his leader will give him a watermelon candy. The problem is how many candies xiaoxin's leader needs to buy?

Input

This problem has multi test cases. First line contains a single integer T(T≤20) which represents the number of test cases.
For each test case, there is a single line containing a string S(1≤length(S)≤1,000).

Output

For each test case, print an integer which is the number of watermelon candies xiaoxin's leader needs to buy after mod 1,000,000,007.

Sample Input

3 aa aabb a

Sample Output

1 2 1

Source

BestCoder Round #77 (div.2)

Recommend

wange2014

code

/**逆元在乘法中的运用*/
#include <bits/stdc++.h>
#define FOR(I,S,T) for(int I=(S);I<=(T);I++)
#define pb push_back
#define mp make_pair
#define eb emplace_backusing namespace std;
typedef long long ll;
const int maxn = 1e4+2;
const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;string s;
ll inv_fac[maxn];
ll fac[maxn];
ll fp(ll x, ll p){if (p == 0) return 1 % mod;ll t = fp(x, p / 2) % mod;if (p & 1) return t * t % mod * x %mod;return t * t % mod;
}
int cnt[maxn];
vector <int> tmp;
int main()
{int T;cin >> T;fac[0] = 1;FOR(i,1,1000) fac[i] = fac[i - 1] * i % mod;inv_fac[1000] = fp(fac[1000], mod - 2ll);for (int i = 999; i >= 1; i--){inv_fac[i] = inv_fac[i + 1] * (i + 1) % mod;}inv_fac[0] = inv_fac[1];int f = 0;while (T--){tmp.clear();cin >> s;sort(s.begin(), s.end());cnt[0]= 1;f = 0;for (int i = 1; i < s.size(); i++){if (s[i] == s[i-1] ) cnt[i]= cnt[i - 1]+1;else {cnt[i] = 1;if (cnt[i -1] & 1) f++;if (cnt[i-1] & 1) tmp.eb(cnt[i-1]-1);else tmp.eb(cnt[i-1]);}}if (s.size() > 1){if (cnt[s.size() - 1] & 1) f++,tmp.eb(cnt[s.size() - 1] -1);else tmp.eb(cnt[s.size() - 1]);}if (f > 1){puts("0");continue;}ll ans = fac[s.size() / 2];for (int i = 0; i < tmp.size(); i++){ans = ans * inv_fac[tmp[i] / 2] % mod;}cout << ans << endl;}return 0;
}

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逆元阶乘逆元数论 HDU 5651 xiaoxin juju needs help

逆元用于用于对除法运算取模的运算

求解逆元

方法一：拓展欧几里得求逆元（求单个数的逆元）

方法二：费马小定理（求单个数的逆元）

方法三：线性求逆元（求一段区间内的所有逆元）

在求解组合数的时候，分子通常会出现某个数的阶乘，下面讲解阶乘逆元

方法一：正推（同时用到 $inv(x) inv(x!)$ ）

方法二：逆推（只用到 $inv(x!)$ ）

例题HDU5651

xiaoxin juju needs help

code

逆元阶乘逆元数论 HDU 5651 xiaoxin juju needs help相关推荐

最新文章

热门文章

逆元 阶乘逆元 数论 HDU 5651 xiaoxin juju needs help

逆元用于用于对除法运算取模的运算

求解逆元

方法一：拓展欧几里得求逆元（求单个数的逆元）

方法二：费马小定理（求单个数的逆元）

方法三：线性求逆元（求一段区间内的所有逆元）

在求解组合数的时候，分子通常会出现某个数的阶乘，下面讲解阶乘逆元

方法一：正推（同时用到）

方法二：逆推（只用到）

例题HDU5651

xiaoxin juju needs help

code

逆元 阶乘逆元 数论 HDU 5651 xiaoxin juju needs help相关推荐

最新文章

热门文章

逆元阶乘逆元数论 HDU 5651 xiaoxin juju needs help

方法一：正推（同时用到 $inv(x) inv(x!)$ ）

方法二：逆推（只用到 $inv(x!)$ ）

逆元阶乘逆元数论 HDU 5651 xiaoxin juju needs help相关推荐