D. Another Problem About Dividing Numbers

【思路】
首先將k=1的情況特判，這種情況只有a%b=0或者b%a=0而且a!=b時才能滿足條件
然後觀察a、b兩個數都可以被除到1，將a,b兩個數進行質因數分解,分解出的質因數個數之和為sum，如果sum>=k則輸出YES，否則輸出NO

/*
 * @Author: Ykui.Chen 
 * @Date: 2021-08-12 17:22:11 
 * @Last Modified by: Ykui.Chen
 * @Last Modified time: 2021-08-12 18:30:34
 */
#include <bits/stdc++.h>
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
inline int read() {
    int X = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') X = (X << 3) + (X << 1) + ch - '0', ch = getchar();
    return X * w;
}
#define ll long long
using namespace std;
const ll maxn = 1e5 + 7;
bool number[maxn + 5];
ll prime[maxn + 5];
ll c = 0;
void isprime() {
    int i, j;
    memset(number, true, sizeof(number));
    for (i = 2; i <= maxn; i++) {
        if (number[i])
            prime[c++] = i;
        for (j = 0; j < c && prime[j] * i <= maxn; j++) {
            number[prime[j] * i] = false;
            if (i % prime[j] == 0)  //保證每個合數只會被它的最小質因數篩去，因此每個數只會被標記一次
                break;
        }
    }
}
ll cal(ll x) {
    ll num = 0;
    for (ll i = 0; i < c; i++) {
        if (x % prime[i] == 0) {
            while (x % prime[i] == 0) {
                num++;
                x /= prime[i];
            }
        }
    }
    if (x > 1)
        num++;
    return num;
}
int main() {
    IOS
    isprime();
    ll t;
  cin>>t;
    while (t--) {
        ll a, b, k;
        cin>>a>>b>>k;
        ll num1 = 0, num2 = 0;
        num1 = cal(a);
        num2 = cal(b);
        ll sum = num1 + num2;
        if (k == 1) {
            if (a % b == 0 || b % a == 0) {
                if (a != b) {
                    puts("YES");
                } else
                    puts("NO");
            } else
                puts("NO");
        } else {
            if (k <= sum) {
                puts("YES");
            } else
                puts("NO");
        }
    }
    return 0;
}
 

這裡有一個困擾了我很久的點，就是要先用素數篩，篩出質數，在用到質因數分解中，否則會超時

F. Interesting Function

【思路】
將l到r之間的數量轉換為[0,r]-[0,l]
計算每一位的貢獻

#include <bits/stdc++.h>
#define int long long
using namespace std;

int cal(int x) {
    int ans = 0;
    while (x) {
        ans += x;
        x /= 10;
    }
    return ans;
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        int l, r;
        cin >> l >> r;
        int ans = cal(r) - cal(l);
        cout << ans << endl;
    }
    return 0;
}