Bela

Young Mirko is a smart, but mischievous boy who often wanders around parks looking for new ideas. This time he’s come across pensioners playing the card game Belote. They’ve invited him to help them determine the total number of points in a game.

Each card can be uniquely determined by its number and suit. A set of four cards is called a hand. At the beginning of a game one suit that “trumps” any other is chosen, and it is called the dominant suit. The number of points in a game is equal to the sum of values of each card from each hand in the game. Mirko has noticed that the pensioners have played NN hands and that suit BB was the dominant suit.

The value of each card depends on its number and whether its suit is dominant, and is given in Table 1.

Number	Value
	Dominant	Not dominant
A	1111	1111
K	44	44
Q	33	33
J	2020	22
T	1010	1010
9	1414	00
8	00	00
7	00	00

Table 1: Scores

Write a programme that will determine and output the number of points in the game.

 Input

The first line contains the number of hands NN (1≤N≤1001≤N≤100) and the value of suit BB (S, H, D, C) from the task. Each of the following 4N4N lines contains the description of a card (the first character is the number of the ii-th card (A, K, Q, J, T, 9, 8, 7), and the second is the suit (S, H, D, C)).

Output

The first and only line of output must contain the number of points from the task.

Sample Input 1Sample Output 1

2 S TH 9C KS QS JS TD AD JH	60

 Sample Input 2 Sample Output 2

4 H AH KH QH JH TH 9H 8H 7H AS KS QS JS TS 9S 8S 7S

题意

给出一个n和一个b，b是钻石牌，然后给4*n的卡牌，计算总积分，如果有b的话则加上面对应的表的第一个数，否则加第二个数

代码

#include<bits/stdc++.h> using namespace std; struct Node {int a1;int a2; } aa[200];int main() {int n;char b;cin >> n >> b;int sum = 0;aa[65].a1 = 11;aa[65].a2 = 11;aa[75].a1 = 4;aa[75].a2 = 4;aa[81].a1 = 3;aa[81].a2 = 3;aa[74].a1 = 20;aa[74].a2 = 2;aa[84].a1 = 10;aa[84].a2 = 10;aa[57].a1 = 14;aa[57].a2 = 0;aa[56].a1 = 0;aa[56].a2 = 0;aa[55].a1 = 0;aa[55].a2 = 0;for(int i = 0; i < 4 * n; i++) {char b1, b2;cin >> b1 >> b2;if(b2 == b) {sum += aa[b1].a1;} elsesum += aa[b1].a2;}cout << sum << endl;return 0; }

 

 

转载于:https://www.cnblogs.com/zhien-aa/p/6279553.html

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