codeforces 546D. Soldier and Number Game

D. Soldier and Number Game time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the second soldier. Then the second one tries to make maximum possible number of rounds. Each round consists of choosing a positive integer x > 1, such that n is divisible by x and replacing n with n / x. When n becomes equal to 1 and there is no more possible valid moves the game is over and the score of the second soldier is equal to the number of rounds he performed.

To make the game more interesting, first soldier chooses n of form a! / b! for some positive integer a and b (a ≥ b). Here by k! we denote the factorial of k that is defined as a product of all positive integers not large than k.

What is the maximum possible score of the second soldier?

Input

First line of input consists of single integer t (1 ≤ t ≤ 1 000 000) denoting number of games soldiers play.

Then follow t lines, each contains pair of integers a and b (1 ≤ b ≤ a ≤ 5 000 000) defining the value of n for a game.

Output

For each game output a maximum score that the second soldier can get.

Examples input 2 3 1 6 3 output 2 5 n=a!/b!,(a>b),

x为质因数时  n/x 至n==1的次数最多

打表a的质因数个数-b的质因数个数

代码：

#include<cstdio> #include<cstring> #define M 5000000+10 int dv[M]; void init()<span style="color:#006600;">//质因数打表</span> { memset(dv,0,sizeof(dv)); for(int i=2;i<M;i++) { if(!dv[i]) { for(int j=i;j<M;j+=i)<span style="color:#006600;">//j的质因数i的个数</span> { int tmp=j; while(tmp%i==0) { dv[j]++; tmp=tmp/i; } } } } for(int i=2;i<M;i++) { dv[i]+=dv[i-1]; } } int main() { init(); int t,a,b; scanf("%d",&t); while(t--) { scanf("%d%d",&a,&b); printf("%d\n",dv[a]-dv[b]); } return 0; }