uva 10820 send a table

所有可能的数对数量为N×N
所求的(x,y)一定满足x,y互素,否则设公约数为t,(x,y)可由(x/t,y/t)得到
若(x,y)互素,那么(y,x)也互素
若能求出满足1<=x<=N,1<=y<=x的互素的(x,y),那么经过对称即可得到所有的数对(可从N×N的二维图上来理解)

求phi值时注意避免溢出

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using namespace std;

const int MAX_N = 50000;

int N;
int phi[MAX_N + 1];

void (int N) {
for (int i = 1; i <= N; i++) phi[i] = i;
for (int i = 2; i <= N; i++) {

if (phi[i] != i) continue;
for (int j = i; j <= N; j += i) {
// first / i then * (i - 1)
// ensures phi[j] <= j and no overflow
phi[j] = phi[j] / i * (i - 1);
}
}
}

int main() {
euler_function(MAX_N);

while (scanf("%d", &N) && N) {
int res = 0;
for (int i = 1; i <= N; i++) res += phi[i];
res = res * 2 - 1;

printf("%dn", res);
}

return 0;
}