PKU 1730

Perfect Pth Powers

Type:挺BT的数学题

解法：X = 2^p1×3^p2×5^p3....(2,3,5...为素因子)，求p1,p2,p3...的最大公约数

Key:由于题目说是"have magnitude(值) at least 2", "(32-bit)", 所以需要用64位整数，并且要考虑负数的情况，很多人都卡在这上面（是题目叙述不清，还是英文差？）。

#include <stdio.h>
#include <string.h>
#define NL 150000

bool p0[NL];
int pnum[NL/9], np;

void prim()
{
    __int64 i, j;
    memset(p0, 0, sizeof(p0));
    for (i=2; i<NL/2; i++) {
        if (!p0[i]) {
            j = i*i;
            while (j<NL) {
                p0[j] = 1;
                j += i;
            }
        }
    }
    np = 0;
    for (i=2; i<NL; i++) {
        if (!p0[i]) pnum[np++] = (int)i;
    }
}

int gcd(int a, int b)
{
    while (b) {
        int t = a;
        a = b;
        b = t%b;
    }
    return a;
}

int main()
{
    prim();
    int cnt[1000];
    __int64 y;    
    while (scanf("%I64d", &y) != EOF ){
        if (!y) break;
        int i=0;
        bool suc = false, pos = false;
        int fact, c = 0;
        if (y < 0) { y = -y; pos = true; }
        __int64 t = pnum[i];        
        while (t*t <= y) {
            if (y%pnum[i] == 0) {
                suc = true;
                fact = pnum[i];
                cnt[c] = 0;
                while (y>1 && y%fact==0) {
                    y /= fact;
                    cnt[c]++;
                }            
                c++;
                if (y == 1) {                    
                    break;
                }
            }
            i++;
            t = pnum[i];
        }
        int gc = cnt[0];
        for (i=1; i<c; i++)
            gc = gcd(gc, cnt[i]);
        if (suc) {
            if (!pos) printf("%d\n", gc);
            else {
                while (gc>1 && !(gc&1)) gc /= 2;
                printf("%d\n", gc);
            }
        }else printf("1\n");
    }
    return 0;
}