//求一个数的质因子,将质因子储存在a[],并返回质因子的个数
int init(ll n) {
ll i;
idx = 0;
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
a[idx++] = i;
while (n % i == 0)
n = n / i;
}
}
if (n > 1)//这里要记得
a[idx++] = n;
return idx - 1;
}
//求[1,m]内与n互质的个数,用队列数组实现容斥原理
ll fid(ll m) {
ll que[10000], k, t = 0, sum = 0;
que[t++] = -1;
for (ll i = 0; i < idx; i++) {
k = t;
for (ll j = 0; j < k; j++)
que[t++] = que[j] * a[i] * (-1);
}
for (ll i = 1; i < t; i++)
sum = sum + m / que[i];
return sum;
}
题目:
代码
ll a[1000], idx;
//求一个数的质因子,将质因子储存在a[],并返回质因子的个数
int init(ll n) {
ll i;
idx = 0;
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
a[idx++] = i;
while (n % i == 0)
n = n / i;
}
}
if (n > 1)//这里要记得
a[idx++] = n;
return idx - 1;
}
//求[1,m]内与n互质的个数,用队列数组实现容斥原理
ll fid(ll m) {
ll que[10000], k, t = 0, sum = 0;
que[t++] = -1;
for (ll i = 0; i < idx; i++) {
k = t;
for (ll j = 0; j < k; j++)
que[t++] = que[j] * a[i] * (-1);
}
for (ll i = 1; i < t; i++)
sum = sum + m / que[i];
return sum;
}
int main() {
ll T, x, y, n, sum;
while (~scanf("%lld", &T)) {
for (ll i = 1; i <= T; i++) {
scanf("%lld%lld%lld", &x, &y, &n);
init(n);
sum = y - fid(y) - (x - 1 - fid(x - 1));
printf("Case #%lld: ", i);
printf("%lld\n", sum);
}
}
return 0;
}
一次求出1~n的所有数的质因子,适用于题目中给的n个数比较多的,但是n不是很大的。
int visited[100010];
vector<int> a[100010];
void init() {
int i, j;
for (i = 0; i < 100010; i++)
a[i].clear();//vector的清空
memset(visited, 0, sizeof(visited));
for (i = 2; i <= 100000; i++) {
if (visited[i] == 0) { //i是素数这是可以保证的
a[i].push_back(i);
for (j = i + i; j <= 100000; j += i) {
visited[j] = 1;
a[j].push_back(i);
}
}
}
}
int main() {
int i, j;
init();
for (i = 0; i <= 50; i++) {
printf("%d:", i);
for (j = 0; j < a[i].size(); j++)
printf("%d ", a[i][j]);
printf("\n");
}
return 0;
}