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Facebook Hacker Cup 2015 Round 1 Homework（附带测试数据）

来源：程序员人生发布时间：2015-01-26 09:01:48 阅读次数：3136次

题目描写：

Homework10 points

Your first-grade math teacher, Mr. Book, has just introduced you to an amazing new concept ― primes! According to your notes, a prime is a positive integer greater than 1 that is divisible by only 1 and itself.

Primes seem fun, but without giving you and your 6-year-old colleagues time to consider their implications, he's promptly gone on to define another term: primacity. He explains that the primacity of an integer is the number of distinct primes which divide it. For example, the primacity of 12 is 2 (as it's divisible by primes 2 and 3), the primacity of 550 is 3 (as it's divisible by primes 2, 5, and 11), and the primacity of 7 is 1 (as the only prime it's divisible by is 7).

Following his lesson, Mr. Book has given you homework with some rather mean questions of the following form: Given 3 integers A, B, and K, how many integers in the inclusive range [A, B] have a primacity of exactly K?

Mr. Book probably expects his little homework assignment to take you and your classmates the rest of the year to complete, giving him time to slack off and nap during the remaining math classes. However, you want to learn more things from him instead! Can you use the skills you've learned in your first-grade computer science classes to finish Mr. Book's homework before tomorrow's math class?

Input

Input begins with an integer T, the number of homework questions. For each question, there is one line containing 3 space-separated integers: A, B, and K.

Output

For the ith question, print a line containing "Case #i: " followed by the number of integers in the inclusive range [A, B] with a primacity of K.

Constraints

1 ≤ T ≤ 100
2 ≤ A ≤ B ≤ 10⁷
1 ≤ K ≤ 10⁹

Explanation of Sample

In the first test case, the numbers in the inclusive range [5, 15] with primacity 2 are 6, 10, 12, 14, and 15. All other numbers in this range have primacity 1.

Example input ・

Example output ・

5 5 15 2 2 10 1 24 42 3 1000000 1000000 1 1000000 1000000 2

Case #1: 5 Case #2: 7 Case #3: 2 Case #4: 0 Case #5: 1

解题思路：

这个题很水，打个表就过了。这个表记录的是某个数有多少个质因数。表的处理方式类似于筛素数。

题目代码：

#include <set> #include <map> #include <queue> #include <math.h> #include <vector> #include <string> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <cctype> #include <algorithm> #define eps 1e⑴0 #define pi acos(⑴.0) #define inf 107374182 #define inf64 1152921504606846976 #define lc l,m,tr<<1 #define rc m + 1,r,tr<<1|1 #define zero(a) fabs(a)<eps #define iabs(x) ((x) > 0 ? (x) : -(x)) #define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A)))) #define clearall(A, X) memset(A, X, sizeof(A)) #define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE)) #define memcopyall(A, X) memcpy(A , X ,sizeof(X)) #define max( x, y ) ( ((x) > (y)) ? (x) : (y) ) #define min( x, y ) ( ((x) < (y)) ? (x) : (y) ) using namespace std; const int M = 10000007; int f[M]; void pri() { int t = 0; for(int i = 2; i <= M; i++) { if(!f[i]) { f[i]++; for(int j=2;i*j<=M;j++) { f[i*j]++; } } //printf("%d %d ",f[i] ,i); } } int main() { //freopen("data.txt","w",stdout); pri(); int t,case1=1; while(scanf("%d",&t)!=EOF) { while(t--) { int a,b,k,ans; scanf("%d%d%d",&a,&b,&k); ans=0; for(int i=a;i<=b;i++) { if(f[i]==k)ans++; } printf("Case #%d: %d ",case1++,ans); } } return 0; }

题目终究测试数据：

链接: http://pan.baidu.com/s/1yGmqa

密码: c5t1

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