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

bool isPrime(int n) {
    if (n <= 1) return false; // 1不是素数
    int sqr = (int)sqrt(1.0 * n);
    for (int i = 2; i <= sqr; i++)
        if (n % i == 0) return false;
    return true;
}

int reverseN(int n, int d) {
    int arr[100] = {0}, len = 0;
    while (n) {
        arr[len++] = n % d;
        n /= d;
    }
    int res = 0;
    for (int i = 0; i < len; i++) {
        res = res * d + arr[i];
    }
    return res;
}

int main() {
    int n, d;
    while (scanf("%d", &n) != EOF && n > 0) {
        scanf("%d", &d);
        if (!isPrime(n)) {
            printf("No\n");
            continue;
        }
        n = reverseN(n, d);
        printf("%s", isPrime(n) ? "Yes\n" : "No\n");
    }

    return 0;
}


73 10
23 2
23 10
-2

Yes
Yes
No

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

int main() {
    char c[4] = {"WTL"};
    double res = 1.0;
    for (int i = 0; i < 3; i++) {
        double maxVal = 0.0;
        int maxChar = 0;
        for (int j = 0; j < 3; j++) {
            double temp;
            scanf("%lf", &temp);
            if (temp > maxVal) {
                maxVal = temp;
                maxChar = j;
            }
        }
        res *= maxVal;
        printf("%c ", c[maxChar]);
    }
    printf("%.2f", (res * 0.65 - 1) * 2);

    return 0;
}


1.1 2.5 1.7
1.2 3.1 1.6
4.1 1.2 1.1

T T W 39.31

发表于

题目大意：
求2个多项式A*B后的结果。

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

int main() {
    int k1, k2, cnt = 0, exp;
    double coe, arr[1001] = {0.0}, ans[2001] = {0.0};
    scanf("%d", &k1);
    for (int i = 0; i < k1; i++) {
        scanf("%d %lf", &exp, &coe);
        arr[exp] += coe;
    }
    scanf("%d", &k2);
    for (int i = 0; i < k2; i++) {
        scanf("%d %lf", &exp, &coe);
        for (int j = 0; j <= 1000; j++) {
            ans[j + exp] += arr[j] * coe;
        }
    }
    for (int i = 2000; i >= 0; i--) {
        if (ans[i]) cnt++;
    }
    printf("%d", cnt);
    for (int i = 2000; i >= 0; i--) {
        if (ans[i])
            printf(" %d %.1f", i, ans[i]);
    }

    return 0;
}


2 1 2.4 0 3.2
2 2 1.5 1 0.5

3 3 3.6 2 6.0 1 1.6