Application of the unique decomposition theorem: The second problem of NOIP2009 raising group Hankson [unique decomposition theorem | violence]

Input/output format

Input format:

The first behavior is a positive integer n, which indicates that there are n groups of input data. The next n rows are each

Line a set of input data, a0,a1,b0,b1 four positive integers, separated by a space between each two integers. Input

The data guarantees that the A0 can be divisible by A1, B1 divisible by B0.

Output format:

Output file Son.out total n rows. The output of each set of input data is one row, which is an integer.

For each set of data: if there is no such x, output 0;

If there is such an X, please output the number of x that satisfies the condition;

Input/Output sample

Input Sample # #:

2

Sample # # of output:

6
2

"description"

The first set of input data, X can be 9, 18, 36, 72, 144, 288, a total of 6.

The second set of input data, X can be 48, 1776, a total of 2.

"Data Range"

For 50% of data, 1≤a0,a1,b0,b1≤10000 and n≤100 are guaranteed.

For 100% of data, 1≤a0,a1,b0,b1≤2,000,000,000 and n≤2000 are guaranteed.

Resolution

1. Decompose the A0,A1,B0,B1 on-line and then judge

2. Brute Force enumeration B1 + various optimizations

Note Enumeration to sqrt (B1) instead of sqrt (B1) +1

//Unique decomposition theorem#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <    Cmath> #include <queue>using namespace std;const int N=50005;inline int read () {char c=getchar (); int x=0,f=1; while (c< ' 0 ' | |    C> ' 9 ') {if (c== '-') f=-1; C=getchar ();}    while (c>= ' 0 ' &&c<= ' 9 ') {x=x*10+c-' 0 '; C=getchar ();} return x*f;}    int n,a0,a1,b0,b1,a,b,ans;int vis[n],p[n],cnt=0;void era (int n) {int m=sqrt (n) +1;    for (int i=2;i<=m;i++) if (!vis[i]) for (int j=i+i;j<=n;j+=i) vis[j]=1; for (int i=2;i<=n;i++) if (!vis[i]) p[++cnt]=i;}    void solve (int p) {int ea0=0,ea1=0,eb0=0,eb1=0;    while (a0%p==0) ea0++,a0/=p;    while (a1%p==0) ea1++,a1/=p;    while (b0%p==0) eb0++,b0/=p;    while (b1%p==0) eb1++,b1/=p;        if (EA0==EA1&AMP;&AMP;EB0==EB1) {if (ea1<=eb1) ans*=eb1-ea1+1;    else ans=0; }else if (ea0!=ea1&&eb0!=eb1&&ea1!=eb1) ans=0;}    int main (int argc, const char * argv[]) {era (50000);    N=read ();while (n--) {a0=read (); A1=read (); B0=read (); B1=read ();        Ans=1;        for (int i=1;i<=cnt;i++) solve (p[i]);        if (b1!=1) solve (B1);    printf ("%d\n", ans); } return 0;}

//Violence #include <iostream> #include <cstdio> #include <cstring> #include <algorithm># Include <cmath> #include <queue>using namespace std;const int N=50005;inline int read () {    char C=getchar ( ); int x=0,f=1;    while (c< ' 0 ' | | C> ' 9 ') {if (c== '-') f=-1; C=getchar ();}    while (c>= ' 0 ' &&c<= ' 9 ') {x=x*10+c-' 0 '; C=getchar ();}    return x*f;} int n,a0,a1,b0,b1,a,b;inline int gcd (int a,int b) {return b==0?a:gcd (b,a%b);} inline int cal (int i) {    if (I%A1) return 0;    else return gcd (i/a1,a) ==1&&gcd (b,b1/i) ==1;} int main (int argc, const char * argv[]) {    n=read ();    while (n--) {        a0=read (); A1=read (); B0=read (); B1=read ();        int Ans=0,m=sqrt (B1);        a=a0/a1,b=b1/b0;        for (int i=1;i<=m;i++) if (b1%i==0) {            ans+=cal (i);            if (I*I!=B1) ans+=cal (b1/i);        }        printf ("%d\n", ans);    }    return 0;}

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Application of the unique decomposition theorem: The second problem of NOIP2009 raising group Hankson [unique decomposition theorem | violence]