Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 351 Accepted Submission(s): 219
Problem Description
In the last semester, Rikka joined the badminton club.
There are n possible registration status which will make the activity fail.
There are n possible registration status which will make the activity fail.
Input
The first line contains a single number ).
Output
For each testcase, output a single line with a single integer, the answer modulo 998244353.
Sample Input
3
1 1 1 1
2 2 2 2
3 4 5 6
Sample Output
12
84
2904
Source
Recommend
题意：没有球没有拍，有拍，有球，有拍又有球的人分别为a,b,c,d，每个人都可以参加比赛，组织一场比赛至少要两个球拍和一个球，问不能组织成功比赛的可能性？
分析：不能成功组织比赛的情况为：
不考虑d：只有拍：2^a*2^b
可能有拍和球：2^a*2^c*(1+b)
减去重复的情况没有拍：2^a
考虑d：只能有一个d：2^a*2^c*d
所以总的不重复情况：2^a*2^b+2^a+2^a*2^c*d-2^a*2^c*(1+b)
参考博客：https://blog.csdn.net/qq_41037114/article/details/81876518
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll qow( ll a, ll b ) {
ll ans = 1;
while(b) {
if(b&1) {
ans = ans*a%mod;
}
a = a*a%mod;
b /= 2;
}
return ans;
}
int main() {
ios::sync_with_stdio(0);
ll T;
cin >> T;
while( T -- ) {
ll a, b, c, d;
cin >> a >> b >> c >> d;
ll ans = qow(2,a)*qow(2,b)%mod;
ans = (ans+((qow(2,a+c)-qow(2,a)+mod)%mod*(1+b))%mod)%mod;
ans = (ans+qow(2,a+c)*d)%mod;
cout << ans << endl;
}
return 0;
}