UVa 10892 - LCM Cardinality （因式分解 组合数学）

Problem F
LCM Cardinality
Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

 

A pair of numbers has a unique LCM but a single number can be the LCM of more than one possible pairs. For example 12 is the LCMof (1, 12), (2, 12), (3,4) etc. For a given positive integer N, the number of different integer pairs with LCM is equal to N can be called theLCM cardinality of that number N. In this problem your job is to find out the LCM cardinality of a number.

 

Input

The input file contains at most 101 lines of inputs. Each line contains an integer N (0<N<=2*10⁹). Input is terminated by a line containing a single zero. This line should not be processed.

 

Output

For each line of input except the last one produce one line of output. This line contains two integers N and C. Here N is the input number and C is its cardinality. These two numbers are separated by a single space.

 

Sample Input                             Output for Sample Input

2

12

24

101101291

0

2 2

12 8

24 11

101101291 5

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Problem setter: Shahriar Manzoor

Special Thanks: Derek Kisman

 

题意：

输入正整数ｎ，统计有多少对正整数a<=b，满足lcm(a, b) = n。

最开始想的方法想到最后一点的时候没有想出来。

对ｎ进行质因式分解：

n = p1^r1 * p2^r2 * ... * px^rx

设：

a = p1^a1 * p2^a2 * ... * px^ax

b = p1^b1 * p2^b2 * ... * px^bx

根据lcm的定义可以知道　ri = max(ai, bi)

假设不考虑a<=b这个条件　则

对于pi，肯定有一个取ri，另外一个随便取有ri+1，所以有(ri+1)*2-1（因为有一种a==b算了两次）=ri*2+1种方案

而ａｂ没有限制所以各个pi之间方案数独立，所以各个pi之间相乘即可

现在考虑a<=b

因为(a1, b1)和(b1, a1)重复了　而恰好包含了a==b的情况（即所有因子ａｂ都取最大）

则 ans = (ans + 1) / 2

另外一种暴力求法

把ｎ的所有因子就出来，然后枚举两个因子　看他们的ｌｃｍ是否等于ｎ，如果是方案数＋１

先挖个坑，另外两个题

NOIP2001普及组

HDU 4497

坑填上了

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 100000 + 20;int vis[maxn], prim[maxn];int A[maxn];void sieve(int n) {    int m = (int) sqrt(n+0.5);    memset(vis, 0, sizeof(vis));    for(int i=2; i<=m; i++) if(!vis[i]) {        for(int j=i*i; j<=n; j+=i) {            vis[j] = 1;        }    }}int getPrimeTable(int n) {    sieve(n);    int c = 0;    for(int i=2; i<=n; i++) if(!vis[i])        prim[c++] = i;    return c;}int main() {    int n;    int primNum = getPrimeTable(100000);    while(scanf("%d", &n) != EOF && n) {        int nn = n;        memset(A, 0, sizeof(A));        for(int i=0; i<primNum && n>1; i++) {            while(n % prim[i] == 0) {                A[i]++;                n /= prim[i];            }        }        LL ans = 1;        for(int i=0; i<primNum; i++) if(A[i]) {            ans *= (2 * A[i] + 1);        }        ans = (ans + 1) / 2;        printf("%d %lld\n", nn, ans);    }    return 0;}

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 100000 + 20;int A[maxn];int gcd(int a, int b) {    return b ? gcd(b, a%b) : a;}int lcm(int a, int b) {    return (LL) a * b / gcd(a, b);}int main() {    int n;    while(scanf("%d", &n) != EOF && n) {        memset(A, 0, sizeof(A));        int num = 0;        int m = (int) sqrt(n+0.5);        for(int i=1; i<=m; i++) if(n % i == 0) {            A[num++] = i;            A[num++] = n / i;        }        if(A[num-1] == A[num-2]) num--;        int ans = 0;        for(int i=0; i<num; i++) {            for(int j=i; j<num; j++) if(lcm(A[i], A[j]) == n) {                ans ++;            }        }        printf("%d %d\n", n, ans);    }    return 0;}