Bzoj2749 [haoi2012] Aliens

Number theory is a good question !!!

First, the formula given by the question, for prime number x> 2, Phi (x) will become a lot 2... while PHI (2) = 1

In YY, we can find the number of 2 resulting from the decomposition of each prime number, and the number of 2 is ans.

So we make f [I] to indicate that I is decomposed into several 2: This process is similar to the prime screening method.

I is a prime number, F [I] = f [I-1]; otherwise, f [I * prime [J] = f [I] + F [prime [J] (<-- this is the legendary pure O (n) linear screening)

In addition, if n is an odd number at the beginning of a period, ANS must + 1 because the first step is required to change to 2.

 1 /************************************************************** 2     Problem: 2749 3     User: rausen 4     Language: C++ 5     Result: Accepted 6     Time:256 ms 7     Memory:1704 kb 8 ****************************************************************/ 9  10 #include <cstdio>11 #include <cmath>12 #include <algorithm>13  14 using namespace std;15 typedef long long LL;16 const int Maxn = 100005;17 LL f[Maxn + 10], ans, del;18 int T, P, Q, m, p[5000], cnt;19 bool x[Maxn + 10];20  21 void pre_work(){22     int t;23     f[1] = 1;24     for (int i = 2; i <= Maxn; ++i){25         if (!x[i])26             p[++cnt] = i, f[i] = f[i - 1];27         for (int j = 1; j <= cnt && i * p[j] <= Maxn; ++j){28             t = i * p[j];29             x[t] = 1;30             f[t] = f[i] + f[p[j]];31             if (!(i % p[j])) break;32         }33     }34 }35  36 int main(){ 37     scanf("%d", &T);38     pre_work();39     while (T--){40         scanf("%d", &m);41         ans = 0, del = 1;42         for (int i = 1; i <= m; ++i){43             scanf("%d%d", &P, &Q);44             if (P == 2) del = 0;45             ans += f[P] * Q;46         }47         printf("%lld\n", ans + del);48     }49     return 0;50 }

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Bzoj2749 [haoi2012] Aliens