Codeforces Round #328 (Div. 2) 592ABC题解

题目链接:

A: 给出8*8的一个棋盘, W为A的棋子, B为B的棋子. 每个棋子只能向上或者向下走, 且不能覆盖或者跳跃棋子, 问你最终谁获胜.

模拟题, 遍历每一列, 记录最考上的W以及最靠下的B的位置.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 10;
char map[MAXN][MAXN];
int a = MAXN, b = MAXN;
int main(int argc, char const *argv[])
{
    for(int i = 0; i < 8; ++i)
        scanf("%s", map[i]);
    for(int j = 0; j < 8; ++j) {
        bool flag = false;
        for(int i = 0; i < 8; ++i) {
            if(map[i][j] == 'W' && !flag) a = min(a, i);
            if(map[i][j] == 'B') flag = true;
        }
        flag = false;
        for(int i = 7; i >= 0; --i) {
            if(map[i][j] == 'B' && !flag) b = min(b, 7 - i);
            if(map[i][j] == 'W') flag = true;
        }
    }
    if(a <= b) printf("A\n");
    else printf("B\n");
    return 0;
}

B: 给出n边形的n, 问可以把此n边形分为几块.

规律题, 自己画几组后就发现答案为(n - 2) ^ 2.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
typedef long long ll;
int main(int argc, char const *argv[])
{
    ll n;
    scanf("%lld", &n);
    n -= 2;
    printf("%lld\n", n * n);
    return 0;
}

C: 有一个跑道, 终点为1 ~ t - 1, 两个人的步速分别为w, b, 问你选择平等的概率是多少.

数学题, 注意到数据的范围, 将要比较的数据转化为double型运用log()函数就可以解决. 通过gcd, lcm即可解决.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "cmath"
using namespace std;
typedef long long ll;
ll t, w, b;
ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}
bool judge(ll a, ll b, ll c)
{
    if(log(1.0 * a) + log(1.0 * b) - log(1.0 * gcd(a, b)) > log(1.0 * c)) return true;
    return false;
}
int main(int argc, char const *argv[])
{
    scanf("%lld%lld%lld", &t, &w, &b);
    if(w == b) {
        printf("1/1\n");
        return 0;
    }
    ll x, y = t;
    if(judge(w, b, t)) x = min(w - 1, min(b - 1, t));
    else {
        ll l = lcm(w, b), g = t / l;
        if(l * g + min(w, b) <= t) x = (g + 1) * min(w, b) - 1;
        else x = g * (min(w, b) - l) + t;
    }
    printf("%lld/%lld\n", x / gcd(x, y), y / gcd(x, y));
    return 0;
}