Light oj 1197 - Help Hanzo

1197 - Help Hanzo

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Time Limit: 2 second(s) Memory Limit: 32 MB

Amakusa, the evil spiritual leader has captured the beautiful princess Nakururu. The reason behind this is he had a little problem with Hanzo Hattori, the best ninja and the love of Nakururu. After hearing the news Hanzo got extremely angry. But he is clever
and smart, so, he kept himself cool and made a plan to face Amakusa.

Before reaching Amakusa's castle, Hanzo has to pass some territories. The territories are numbered as a, a+1, a+2, a+3 ... b. But not all the territories are safe for Hanzo because there can be other fighters waiting for him. Actually he is not afraid of them,
but as he is facing Amakusa, he has to save his stamina as much as possible.

He calculated that the territories which are primes are safe for him. Now given a and b he needs to know how many territories are safe for him. But he is busy with other plans, so he hired you to solve this small problem!

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a line containing two integers a and b (1 ≤ a ≤ b < 231, b - a ≤ 100000).

Output

For each case, print the case number and the number of safe territories.

Sample Input

Output for Sample Input

3

2 36

3 73

3 11

Case 1: 11

Case 2: 20

Case 3: 4

Note

A number is said to be prime if it is divisible by exactly two different integers. So, first few primes are 2, 3, 5, 7, 11, 13, 17, ...

用小素数来筛选掉大的和数，，，，，注意数组大小和long long........wrong了几十次，和别人的代码一行一行的对才发现是该用long long 的锅

代码：

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
#define MA 70005   //65536，，数组的大小
#define LL long long
int pr[MA],kk=0,t;
bool shu[101000],prim[MA];
LL a,b;
int main()
{
long long i,j;
memset(prim,true,sizeof(prim));
for (i=2;i<MA;i++)
if (prim[i])
{
pr[kk++]=i;
for (j=i+i;j<MA;j+=i)
prim[j]=false;
}scanf("%d",&t);
for (int ca=1;ca<=t;ca++)
{
scanf("%lld%lld",&a,&b);
memset(shu,true,sizeof(shu));
if (a<2) a=2;
long long sp,pp=b-a,s=0;
for (i=0;i<kk;i++)
{
sp=pr[i];
j=sp*(a/sp);
if (j<a)
j+=sp;
if (j<sp*sp)//j和sp都要用long long,,,,
j=sp*sp;
for (;j<=b;j+=sp)
shu[j-a]=false;
}
for (i=0;i<=pp;i++)
if (shu[i])
s++;
printf("Case %d: %lld\n",ca,s);
}
return 0;
}