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Description
Input
Output
Sample Input
3 1 100 2 2 3 19
Sample Output
Case #1: 14 Case #2: 1 Case #3: 4
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int ans=0;
bool p[1000000];
int num[1000000];
void prime()
{
int i,j;
memset(p,0,sizeof(p));
p[1]=1;
for(int i=2; i<=1000000; i++)
{
if(!p[i])
{
for(j=2; i*j<=1000000; j++)
{
p[i*j]=1;
}
}
}
}
void biao()
{
int a,b,c,d,e,f,g;
memset(num,0,sizeof(num));
for(int i=2; i<=1000000; i++)
{
if(!p[i])
{
a=i/1000000;
b=i%1000000/100000;
c=i%100000/10000;
d=i%10000/1000;
e=i%1000/100;
f=i%100/10;
g=i%10;
if(!p[a+b+c+d+e+f+g])
ans++;
}
num[i]=ans;
}
}
int main()
{
int T,L,R;
int a,b,c,d,e,f,g;
int ans;
int x=1;
scanf("%d",&T);
prime();
biao();
while(T--)
{
scanf("%d%d",&L,&R);
printf("Case #%d: %d\n",x++,num[R]-num[L-1]);
}
return 0;
}
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原文地址:http://blog.csdn.net/a1967919189/article/details/47731879