杭电多校第九场 hdu6425 Rikka with Badminton 组合数学 思维

Rikka with Badminton

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 students in the badminton club, some of them have rackets, and some of them have balls. Formally, there are a students have neither rackets nor balls, bstudents have only rackets, c students have only balls, and d students have both rackets and balls. (a+b+c+d=n)

This week, the club is going to organize students to play badminton. Each student can choose to take part in or not freely. So there are 2n possible registration status.

To play badminton, there must be at least two students who have rackets and at least one students who have balls. So if there aren't enough balls or rackets, the activity will fail. 

Now, Rikka wants to calculate the number of the status among all 2n possible registration status which will make the activity fail.

 

Input

The first line contains a single number t(1≤t≤103), the number of testcases.

For each testcase, the first line contains four integers a,b,c,d(0≤a,b,c,d≤107,a+b+c+d≥1).

 

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

2018 Multi-University Training Contest 9

 

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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;
}