HDU 4704 Sum (partition principle + Fermat theorem)

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=4704

Test instructions

Given a number n to decompose it, Si represents the number of schemes to split n into the number of I

Seek sum (SI) 1<=i<=n;

Analysis:

Partition principle, n a stick, n-1 a seam,

Divided into 1 parts is C (n-1,0);

Divided into 2 parts is C (n-1,1);

Divided into 3 parts is C (n-1,2);

...

Divided into n parts is C (n-1,n-1);

ans = SUM (C (n-1,i)) (0<=i<=n-1)

=2^ (n-1);

Due to modulo and 2 and mod coprime, you can use Fermat theorem to power down

The code is as follows:

#include <iostream> #include <cstring> #include <cstdio>using namespace std;typedef long Long ll;const int mod = 1e9+7;const int maxn = 1e5+10;char A[MAXN]; ll Quick_mod (ll a,ll b,ll m) {    ll ans = 1;    while (b) {        if (b&1) {            ans = ans * a% m;            b--;        }        b>>=1;        A = A * a% m;    }    return ans;} ll Change (char *s, ll m) {    ll ans = 0;    int len = strlen (s);    for (int i = 0; i < len; i++) {        ans = (ans * + a[i]-' 0 ')%m;    }    return ans;} int main () {    while (~scanf ("%s", a)) {        int m = mod-1;        LL n = Change (a,m);        printf ("%i64d\n", Quick_mod (2, (n-1+m)%m,mod));    }    return 0;}



HDU 4704 Sum (partition principle + Fermat theorem)