http://acm.hdu.edu.cn/showproblem.php?pid=5104
找元组数量,满足p1<=p2<=p3且p1+p2+p3=n且都是素数
不用素数打表都能过,数据弱的一比
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <sstream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
const int inf = 0x7fffffff;
const int maxn = 1e5+5;
int p[maxn],pr[maxn],cnt = 0;
int n;
void init()
{
clr0(p);
for(int i = 2;i < maxn;++i){
if(!p[i]){
for(int j = i + i;j < maxn;j+=i)
p[j] = 1;
}
}
p[0] = 1;
for(int i = 2;i < maxn;++i)
if(!p[i])
pr[cnt++] = i;
}
int main()
{
init();//cout<<cnt;
while(~RD(n)){
LL ans = 0;
for(int i = 0;i < cnt;++i){
if(n < pr[i]*3)
break;
for(int j = i;j < cnt;++j){
int k = n - pr[i] - pr[j];
if(pr[j] > k)
break;
//printf("%d %d %d \n",pr[i],pr[j],k);
if(!p[k])
ans++;
}
}
printf("%I64d\n",ans);
}
return 0;
}