Bzoj 2005 [Noi2010] energy harvesting (tolerance)

[Noi2010] Energy harvesting Time Limit: ten Sec Memory Limit:552 MB
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The building has a rectangular land, he planted an energy plant on the ground, which can collect the energy of sunlight. After the plants collected energy, the building then used an energy pooling machine to bring together the energy collected by the plants. The plants in the building are very neatly planted, there are n columns, each of which has m, and the plant's spacing is the same, so for each plant, the building can be represented by a coordinate (x, y), where the range of x is 1 to n, that is in column X, the range of Y is 1 to M, and that is the nth tree in column X. Because the energy collection machine is large and inconvenient to move, the building puts it on a corner with coordinates exactly (0, 0). The energy pooling machine has a certain energy loss during the collection process. If a plant is connected to an energy collection machine with a K plant on its line, the loss of energy is 2k + 1. For example, when the energy collection machine collects plants with coordinates of (2, 4), there is a 3 loss of energy due to the presence of a plant (1, 2) on the connecting segment. Note that if a plant has no plants on the line segment connected to the energy pooling machine, the energy loss is 1. Now we have to calculate the total energy loss. An example of energy harvesting is given below, where n = 5,m = 4, a total of 20 plants, showing the energy loss generated by the energy collection machine on each plant. In this example, a total of 36 of the energy loss is generated.

Input

Contains only one row, which is two integers n and M.

Output

Contains only an integer that represents the total energy loss generated.

Sample Input "Sample input 1"
5 4

"Sample Input 2"
3 4
Sample Output "Sample Output 1"
36

"Sample Output 2"
20

"Data size and conventions"
For 10% data: 1≤n, m≤10;

For 50% data: 1≤n, m≤100;

For 80% data: 1≤n, m≤1000;

For 90% data: 1≤n, m≤10,000;

For 100% data: 1≤n, m≤100,000.

Title Link: http://www.lydsy.com/JudgeOnline/problem.php?id=2005

Title Analysis: First of all it is not difficult to find the number of plants between the points (x, y) and (0,0) is gcd (x, y)-1, so the subject is actually Ōi (1-n) Σj (1-M) [2 * (GCD (i, J)-1)-1], simplifying the 2 *σi (1-n) σj (1-m) CD (I, j)-N * m, now the problem is how to quickly find Ōi (1-n) σj (1-m) gcd (i, J), can be used Möbius inversion, but the direct nlogn of the allowance can be, Cnt[i] recorded is greatest common divisor for I of the two-tuple number, first ( n/i) * (m/i) is all the number of the two-tuple group I as the Convention, then take Cnt[i] minus all the cnt[j] (J is a multiple of I), the rest is all the two-tuple number with I as the greatest common divisor, notice here enumeration approximate when the reverse order, because we want to use small to reduce the size, To ensure that the big has been calculated, and then according to the formula calculation can be, notice to use a long long

#include <cstdio> #include <cstring> #include <algorithm> #define LL long longusing namespace Std;int Const MAX = 1e5 + 5;ll cnt[max];int main () {    ll ans = 0;    memset (CNT, 0, sizeof (CNT));    ll N, m;    scanf ("%lld%lld", &n, &m);    if (n < m)        swap (n, m);    for (int i = n; I >= 1; i--)    {        Cnt[i] = (ll) (n/i) * (m/i);        for (int j = i * 2; J <= N; j + = i)            cnt[i]-= cnt[j];        Ans + = i * cnt[i];    }    printf ("%lld\n", 2 * ans-n * m);}

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Bzoj 2005 [Noi2010] energy harvesting (tolerance)