Introduction to Algorithmic competition 10.1 number of preliminary example code of number theory

10.1 Colossal Fibonacci numbers! UVA11582 idea: Cyclic section + fast Power

#include <cstdio> #include <cstring> #include <algorithm> #define LLU Long Long unsignedusing namespace    std;inline int Qpow (LLu x,llu y, int MOD) {x%=mod;    LLu ans = 1,tem = x;        while (y) {if (y&1) ans = (ans*tem)%mod;        TEM = (TEM * tem)%MOD;    y/=2; }return (int) ans; LLu A[1000000+5];//1 1 2 3 5 8 21inline void print (int n) {for (int i=0;i<n;i++) printf ("%llu%llu\n", A[i], A[i+1]);}    int main () {LLu A, B;    int t;scanf ("%d", &t);        while (t--) {int n;        scanf ("%llu%llu%d", &a,&b,&n); a[0]=0%n;        A[1]=1%n;int M=n*n;            for (int i=2;i<=m;i++) {a[i]= (a[i-1]+a[i-2])%n;                if (A[i]==a[1]&&a[i-1]==a[0]) {n=i-1;            Break        }}//print (n);        int ans = QPOW (a,b,n);//printf ("ans =%d\n", ans);    printf ("%llu\n", A[ans]); }}

10.3 Choose and Divide UVA10375 idea: The only number of prime numbers to decompose

#include <cmath> #include <cstdio> #include <cstring> #include <algorithm> #define __int64 Long longusing namespace Std;const int N = 10000+5;inline __int64 C (__int64 n,__int64 m) {}int prime[2000];int Num[2000l];bool J    udge_prime[n]={0};inline int init () {int num = 0;    memset (judge_prime,false,sizeof (judge_prime));            for (int i=2;i<n;i++) {if (!judge_prime[i]) {prime[num++]=i;        for (int j=i*i;j<n;j+=i) judge_prime[j]=true; }}return num;} inline void add_primefactor (int n,int d,int primenum) {for (int. i=0;i<primenum&&n>1;i++) {while (n%prime[i]==0)            {Num[i]+=d;        N/=prime[i]; }}}inline void Add_fun (int n,int d,int primenum) {for (int i=2;i<=n;i++) add_primefactor (i,d,primenum);} void print (int n) {for (int i=0;i<n;i++) printf ("%d-->%d\n", Prime[i],num[i]), GetChar ();} inline void work (int a,int b,int c,int d,int primenum) {   Add_fun (A,1,primenum);    Add_fun (C,-1,primenum);    Add_fun (B,-1,primenum);    Add_fun (D,1,primenum);    Add_fun (A-b,-1,primenum); Add_fun (c-d,1,primenum);}    int main () {int primenum = init ();    int a,b,c,d;        while (scanf ("%d%d%d%d", &a,&b,&c,&d) ==4) {memset (num,0,sizeof (num));        Work (A,b,c,d,primenum);        Double ans = 1;            for (int i=0;i<primenum;i++) {ans *= pow (prime[i],num[i]);        if (Prime[i]>=max (a,c)) break;    } printf ("%.5f\n", ans); }}

10.4 minimun Sum LCM UVA10791 idea: the only decomposition

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath>using namespace std; inline int work (int a,int& x) {    int res = 1;    while (x%a==0) {        res*=a;        X/=a;    } return res;} int main () {    int n,t=0;    while (scanf ("%d", &n) ==1&&n)    {        if (n==1) {            printf ("Case%d:%d\n", ++t,2);            Continue;        }        int k=0;        Long Long ans = (long long) n+1;        int cur = 0;int M=N/2;        for (int i=2;i<=m&&i<=n;i++)        {            if (n%i==0) {                k++;                Cur + = Work (i,n);}        } printf ("cur=%d\n", cur);        if (cur) {            ans = min (ans, k<2?cur+1ll:cur+0ll);        }        printf ("Case%d:%lld\n", ++t,ans);}    }

10.5 GCD XOR UVA12716
Idea: The basic nature of XOR: A^b=c,a^c=b + sieve Prime

#include <cstdio> #include <cmath> #include <cstring> #include <algorithm>using namespace std; const int N = 30000000+5;int a[n];inline void init () {    int x = N;    memset (a,0,sizeof (A));    for (int i=1;i<=x;i++)    {        a[i]+=a[i-1];        for (int j=i*2;j<=x;j+=i)        {            int l= j^i;            if (j>l&& (j-l) ==i)                a[j]++;//printf ("a=%d,b=%d,c=%d\n", J,j^i,i);}}}    int main () {    int t,t=0;init ();    scanf ("%d", &t);    while (t--)    {        int n;        scanf ("%d", &n);        printf ("Case%d:%d\n", ++t,a[n]);}    }

Getting started with algorithmic competition 10.1 number of initial examples code