ural 1049. Brave Balloonists

1049. Brave Balloonists

Time limit: 2.0 second
Memory limit: 64 MB

 

Ten mathematicians are flying on a balloon over the Pacific ocean. When they are crossing the equator they decide to celebrate this event and open a bottle of champagne. Unfortunately, the cork makes a hole in the balloon. Hydrogen is leaking out and the balloon is descending now. Soon it will fall into the ocean and all the balloonists will be eaten by hungry sharks.

But not everything is lost yet. One of the balloonists can sacrifice himself jumping out, so that his friends would live a little longer. Only one problem still exists: who is the one to get out. There is a fair way to solve this problem. First, each of them writes an integer a_i not less than 1 and not more than 10000. Then they calculate the magic number N that is the number of positive divisors of the product a₁*a₂*…*a₁₀. For example, the number of positive integer divisors of 6 is 4 (they are 1,2,3,6). The hero (a mathematician who will be thrown out) is determined according to the last digit of N. Your task is to find this digit.

Input

Input contains ten integer numbers (each number is in separate line).

Output

Output a single digit from 0 to 9 — the last digit of N.

Sample

input	output
1 2 6 1 3 1 1 1 1 1	9

Problem Author: Stanislav Vasilyev
Problem Source: Ural State University collegiate programming contest (25.03.2000)

题目大意：求10个数积因子的个数，输出最后一个数字

思路：

n = a^p1*b^p2*c^p3 ,a b, c 为质数

n因子的个数为 (1+p1)*(1+p2)*(1+p3)

单个分解每个数， 最后用公式。。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>
#define FOR(i, a, b)  for(int i = (a); i <= (b); i++)
#define RE(i, n) FOR(i, 1, n)
#define FORP(i, a, b) for(int i = (a); i >= (b); i--)
#define REP(i, n) for(int i = 0; i <(n); ++i)
#define SZ(x) ((int)(x).size )
#define ALL(x) (x).begin(), (x.end())
#define MSET(a, x) memset(a, x, sizeof(a))
using namespace std;


typedef long long int ll;
typedef pair<int, int> P;
int read() {
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
const double pi=3.14159265358979323846264338327950288L;
const double eps=1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 10005;
const int xi[] = {0, 0, 1, -1};
const int yi[] = {1, -1, 0, 0};

int N, T;
int a[MAXN], cnt, prime[MAXN];
void init() {
    cnt = 0;
    for(int i = 2; i <= 10000; i++) {
        int flag = 1;
        for(int j = 2; j*j <= i; j++) {
            if(i%j == 0) {
                flag = 0;
                break;
            }
        }
        if(flag) prime[++cnt] = i;
    }
}
int main() {
    //freopen("in.txt", "r", stdin);
    init();
    memset(a, 0, sizeof(a));
    int N = 1, M;
    for(int i = 0; i < 10; i++) {
        scanf("%d", &N);
        for(int j = 1; j <= cnt; j++) {
            // printf("%I64d
", N);
            while(N%prime[j] == 0 && N > 1) {
                N /= prime[j];
                a[j]++;
            }
        }
    }
    int s = 1;
    for(int i = 1; i <= cnt; i++) {
        s = s*(a[i]+1)%10;
    }
    printf("%d
", s);
    return 0;
}