B1018. Hammer and scissors

Title Description

　　Everyone should play "a hammer and scissors" game: they both give gesture, as shown in the outcome of the rules

　　Now given the two men clash records, statistics of both wins, flat, negative number, and gives both parties are at the greatest chance of winning what gesture

Input Format

　　The first line gives a positive integer N (≤ 10 ^{. 5} ), i.e., both the number of confrontation. Then N rows, each row gives information confrontation, i.e., the gesture and B sides given simultaneously. C stands for "hammer", the representative J "see", B stands for "cloth", the first letter represents a party, on behalf of the second letter B, the middle of a space

Output Format

　　First, two lines are given A, B wins, flat, negative times, with a space between digits. The third line is given two letters represent A, B wins the highest number of gestures, a space intermediate

If the solution is not unique, the minimum output lexicographical ordering Solutions

SAMPLE INPUT

Sample Output

5 3 2

2 3 5

B B

#include <bits/stdc++.h>
int change(char c){
    if(c == 'B'){
        return 0;
    }
    if(c == 'C'){
        return 1;
    }
    if(c == 'J'){
        return 2;
    }
}
char unchange(int n){
    if(n == 0){
        return 'B';
    }
    if(n == 1){
        return 'C';
    }
    if(n == 2){
        return 'J';
    }
}
int main(int argc, char *argv[]) {
    int n;     
    Scanf ( " % D " , & n-); // Number of rows 
    int times_A [ . 3 ] = { 0 , 0 , 0 }, times_B [ . 3 ] = { 0 , 0 , 0 }; // record the number of times even negative AB 
    char C1, C2;
     int K1, K2;
     int count1 is [ . 3 ] = { 0 , 0 , 0 }, count2 [ . 3 ] = { 0 , 0 , 0 }; // record A, b number of wins
    int MAX1, MAX2, index1,, index2; // record corresponding to the maximum and the maximum array index 
    for ( int I = 0 ; I <n-; I ++ ) {
        getchar();
        Scanf ( " % C% C " , & C1, & C2);
         // gesture into digital 
        K1 = Change (C1);
        K2 = Change (C2);
         // A win 
        IF ((K1 + . 1 )% . 3 == K2) {
            times_A[0]++;
            times_B[2]++;
            count1[k1]++;
        // 平手 
        }else if(k1 == k2){
            times_A[1]++;
            times_B[1]++;
        // 乙赢 
        }else if((k2 + 1)%3 == k1){
            times_A[2]++;
            times_B[0]++;
            count2[k2]++;
        }
    }         
    printf("%d %d %d\n", times_A[0], times_A[1], times_A[2]);    
    printf("%d %d %d\n", times_B[0], times_B[1], times_B[2]);
    max1 = count1[0];
    max2 = count2[0];
    index1 = 0;
    index2 = 0;
    for(int i = 1;i < 3; i++){
        if(max1 < count1[i]){
            max1 = count1[i];
            index1 = i;
        }
        if(max2 < count2[i]){
            max2 = count2[i];
            index2 = i;
        }
    }
    printf("%c %c", unchange(index1), unchange(index2));
    return 0;    
}

The key problem solution

Shengping Fu calculated when the gesture indicating letters in lexicographic order conversion digital computing, B, C, J, are converted to 0,1,2,0 wins wins 2,2 1,1 0 wins, the cycle is determined