cf1561B. Charmed by the Game

Posted 2021-09-11 Jozky86

cf1561B. Charmed by the Game

题意：

两人轮流发球，有两种得分的情况，一个是自己发球的回合得分，叫做"holds"，另一种是在对方发球的回合得分，叫做"breaks"，现在给出比赛结束后两个人的得分，问你两个人总的"breaks"的次数有多少种情况。不知道谁先发球，也不知道谁哪个回合取胜，只知道最后的得分

题解：

我们知道两人得分是x，y，总比赛数量就是x+y,先手发球次数为p=$\lceil \frac{a+b}{2}\rceil$，后手为q=$\lfloor \frac{a+b}{2}\rfloor$
现在也不知道谁先发球，我们可以设Alice先发球，设x为Alice输掉场次为a(0<=a<=p),设Borys输掉场次为b(0<=b<=q)
如果Alice先发球，枚举Alice输的个数a，从0到p，然后有x=(p-a)+y,y=x-(p-a),只要y满足(0<=y<=q),这就是合法的情况，k就是x+y
Borys发球时同理

代码：

// Problem: B. Charmed by the Game
// Contest: Codeforces - Codeforces Round #740 (Div. 2, based on VK Cup 2021 - Final (Engine))
// URL: https://codeforces.com/contest/1561/problem/B
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// Data:2021-08-24 23:07:04
// By Jozky

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCAL
    startTime= clock();
    freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCAL
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
int main()
{
    //rd_test();
    int t;
    read(t);
    while (t--) {
        int x, y;
        read(x, y);
        if ((x + y) % 2) {
            int ans= 2 * (min(x, y) + 1);
            printf("%d\\n", ans);
            int tot= (x + y) / 2 - min(x, y);
            for (int i= 1; i <= ans; i++) {
                printf("%d ", tot);
                tot++;
            }
        }
        else if ((x + y) % 2 == 0) {
            int ans= (min(x, y) + 1);
            printf("%d\\n", ans);
            int tot= (x + y) / 2 - min(x, y);
            for (int i= 1; i <= ans; i++) {
                int ans= (x + y) / 2 - min(x, y) + 2 * (i - 1);
                printf("%d ", ans);
            }
        }
        printf("\\n");
    }
    return 0;
    //Time_test();
}