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=⌈a+b2⌉\lceil \frac{a+b}{2} \rceil⌈2a+b⌉，后手为q=⌊a+b2⌋\lfloor \frac{a+b}{2} \rfloor⌊2a+b⌋
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 LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= 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();
}
)
)
