HDU 4704 Sum (integer and decomposition + fast Power + Fermat theorem Power down)

Test instructions

Give N (1<n<), (S1+S2+S3+...+SN) mod (1e9+7).                         Where Si represents a number of n by the number of I added, such as n=4, then s1=1,s2=3. (End of whole text)

Knowledge Points:

Integer n the kind and decomposition method.

Fermat theorem: P is prime, if p cannot divide a, then a^ (p-1) ≡1 (mod p). The Ma Xiao theorem can be used to reduce the power of primes.

When M is a prime number, (M must be a prime number to use the Fermat theorem)

When a=2. (a=2 is only the condition in question, a can be other values)

MoD m = *//k=

=//==1 for the application of Fermat theorem

For example, set p=7, n=32, ask for 2^32≡x (mod p) value

Because P is a prime number, there must be 2^6≡1 (mod p)

The

2^32%p= (2^[(6*5) +2])%p

=[2^ (6*5) *2^2]%p

=[(2^ (6*5)%p) * (2^2%p)]%p//(A*B)%m=[(a%m) * (b%m)]%m;

=[1* (2^2%p)]%p//2^ (6*5)%p for the application of Fermat theorem

=2^2%p;

Exercises

The problem is equal to the number of decomposition of N. For example, n=x1+x2+x3+. XK is a decomposition that regards XI as being made up of Xi 1, and N is the N 1 component.

The topic is the number of methods that give N 1 groupings (this is not a ball spacer problem similar to combinatorial math). Every two 1 whether to put a baffle, there are put and not two options, altogether n-1 an optional interval. So the total number of methods is.

Because N is too big, not good processing ah.

If the exponent is too large and the m=1e9+7 is found to be a prime number, the Fermat theorem (a^ (p-1)) ≡1 (mod p) can be used to power down.

#include <iostream> #include <cstdio> #include <cstring>using namespace std;typedef long Long ll;const int Mod=1e9+7,n=1e5+5;char A[n]; ll Quick_mod (ll a,ll P)          //Fast power (fast power using two points thought and Qin algorithm) {    ll ans=1;    while (p)    {        if (p&1)            ans=ans*a%mod;        A=a*a%mod;        p>>=1;    }    return ans;} int main () {    while (~scanf ("%s", a))    {        int len=strlen (a);        LL ans=0;        for (int i=0;i<len;i++)        {            ans= (ans*10+a[i]-' 0 ')% (mod-1);        }        ans= (ans-1+mod-1)% (mod-1);        printf ("%lld\n", Quick_mod (2,ans));    }    return 0;}

This problem can also find a loop knot.

Found 2^500000003 = 1 = 2^0, so n= (n-1)%500000003, so 2^ (n-1) = 2^ ((n-1)% (mod-1))%mod; (mod-1=500000003)

Sum

Time limit:1000MS Memory Limit:131072KB 64bit IO Format:%i64d &%i64u

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HDU 4704 Sum (integer and decomposition + fast Power + Fermat theorem Power down)