Project Euler_Problem 193_Few Repeated Digits_欧拉筛+容斥公式

解题思路：暴力搜索

代码：

void solve() {ll i, j,k,x,y,z,p,q,u,v,l,l1;N = 999966663333, NN = 1024;//N = 1000;double a, b, c,d;M.NT.get_prime_Euler(1000000);l = M.NT.pcnt;for (i = 1; i <= l; i++) {u = M.NT.prime[i];v = M.NT.prime[i + 1];x = u * u;y = v * v;if (x > N)break;for (j = x; j < y; j=j+u) {if (j > N)break;if (j == x)continue;if (j % v == 0)continue;ans1 = ans1 + j;}for (j = y; j > x; j = j - v) {if (j > N)continue;if (j == y)continue;if (j % u == 0)continue;ans1 = ans1 + j;}}printf("%lld\n",ans1);
}