Nyoj 411 Friends Number (number theory-factor and)

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Test instructions

Friends number time limit: ms | Memory limit: 65535 KB Difficulty: 2

Describe

Paula and Tai are couple. There is many stories between them. The day Paula-airplane, Tai send one message to telephone 2200284, then, everything is changing ... (The story in "The Snow Queen"). After a long time, Tai tells Paula, the number and 284 was a couple of friends number, as they is special, all divisor S of 284, and all divisors of 284 ' s sum is 220. Can you find out there is how many couples of friends number less than 10,000. Then, what about 100,000, 200,000 and so on. The task for-is-find out there-many couples of friends number in given closed interval [a, b].

Input

there is several cases.
Each test case contains positive integers a, B (1<= a <= b <=5,000,000).
Proceed to the end of file.

Output

for each test case, output the number of couples in the given range. The output of one test case occupied exactly one line.

Sample input

1 1001 1000

Sample output

01

Tips

6 is a number whose sum of all divisors are 6.6 is not a friend number, these number is called Perfect number.

Source

Liaoning Province 10-year provincial race

Idea: Actually understand test instructions is relatively simple, to find the interval to meet "friend Num" number of pairs, friend Num: Suppose, a, B, if all the factors of a and ==b, and all of the factors and ==a, and meet!

OJ data a bit large, conventional method, kept time-out, looked at the first few people in the code, the basic is to play the table, (embarrassing ~ ~), later a hint, with the array simulation, in fact, can also use the container

, it's more comfortable to write.

Code: (red area emphasis)

 #include <stdio.h> #include <iostream> #include <string.h> #include <algorithm>using namespace STD; #define N 5000001#define CLR (arr, what) memset (arr, what, sizeof (arr)) int a[n], pre[n],last[n],ss=0;void getsum ();        int main () {//getsum ();/for (int i = 2, I <= N; i++) A[i] = 1;//for (int i = 2; I * I <= N; i++)//   for (int j = i; J * I <= N; j + +)//A[i * j] + = i + j; <span style= "color: #ff0000;"    > for (int i = 2; I <= N; i++) a[i] = 1;    for (int i = 2, I * I <= N; i++) for (int j = i + 1; I * j <= N; j + +) A[i * j] + = i + j;    for (int i = 2; I * I <= N; i++) a[i * I] + = i;        for (int i=1; i<=n; i++) {int t=a[i];            if (T > I && t <=N&& a[t] = = i) {pre[ss]=i;            last[ss]=t;        ss++;    }}</span> int m,i,j,n;        while (~SCANF ("%d%d", &m,&n)) {int count = 0; for (i=0; i<ss; i++) if (pre[i]>=m&&last[i]<=n) count++;    printf ("%d\n", count);         } return 0;}

When you want to give up, think of what you persist until now!

Nyoj 411 Friends Number (number theory-factor and)