Thu, 24 Sep 2020 16:54:25 GMT

hdu/6889.cpp

@@ -63,15 +63,17 @@ inline ll fpow(ll a, ll b, ll m)
 }
 constexpr double eps = 1e-8;
 inline int sgn(double x) { return x > eps ? 1 : x < -eps ? -1 : 0; }
+namespace Min25
+{
 const int N = 1000010;
 int prime[N], id1[N], id2[N], flag[N], ncnt, m;
 ll g[N], sum[N], a[N], T, n;
 inline int ID(ll x) { return x <= T ? id1[x] : id2[n / x]; }
 inline ll calc(ll x) { return x * (x + 1) / 2 - 1; }
 inline ll f(ll x) { return x; }
-void init()
+void initPrime()
 {
-    T = sqrt(n + 0.5);
+    T = 1e5 + 50;
     for (int i = 2; i <= T; i++)
     {
         if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
@@ -81,6 +83,11 @@ void init()
             if (i % prime[j] == 0) break;
         }
     }
+}
+void init()
+{
+    m = 0;
+    T = sqrt(n + 0.5);
     for (ll l = 1; l <= n; l = n / (n / l) + 1)
     {
         a[++m] = n / l;
@@ -91,10 +98,14 @@ void init()
         g[m] = calc(a[m]);
     }
     for (int i = 1; i <= ncnt; i++)
-        for (int j = 1; j <= m && (ll)prime[i] * prime[i] <= a[j]; j++)
-            g[j] = g[j] - (ll)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
+        for (int j = 1; j <= m && (ll)prime[i] * prime[i] <= a[j]; j++) g[j] = g[j] - (ll)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
+}
+inline ll solve(ll x)
+{
+    if (x <= 1) return x;
+    return n = x, init(), g[ID(n)];
 }
-
+} // namespace Min25
 ll calc1(ll _n, ll mod)
 {
     __int128 x = _n;
@@ -102,8 +113,7 @@ ll calc1(ll _n, ll mod)
 }
 int main()
 {
-    n = 1e10 + 10;
-    init();
+    Min25::initPrime();
     int T;
     ll __N, mod;
     read(T);
@@ -111,7 +121,7 @@ int main()
     {
         read(__N, mod);
         ll ans = calc1(__N, mod);
-        ans = ans + g[ID(__N + 1)];
+        ans = ans + Min25::solve(__N + 1) % mod;
         printf("%lld\n", ans % mod);
     }
     return 0;