[noip2014]生活大爆炸版石头剪刀布

Posted 2020-12-01 zcr-blog

传送门

题意就不说了，算法也就是个大暴力，把每种情况输赢都预处理出来（打表就好了），注意一些细节就可以了。

上代码：

#include <iostream>
using namespace std;
int n, na, nb;
int a[210], b[210];
int aa, ab;
/*
1:平 
2:赢
3:输 
*/
int cmp(int x, int y) {
    switch (x) {
        case 0:
            switch(y) {
                case 0:
                    return 1;
                case 1:
                    return 3;
                case 2:
                    return 2;
                case 3:
                    return 2;
                case 4:
                    return 3;
            }
        case 1:
            switch(y) {
                case 0:
                    return 2;
                case 1:
                    return 1;
                case 2:
                    return 3;
                case 3:
                    return 2;
                case 4:
                    return 3;
            }
        case 2:
            switch(y) {
                case 0:
                    return 3;
                case 1:
                    return 2;
                case 2:
                    return 1;
                case 3:
                    return 3;
                case 4:
                    return 2;
            }
        case 3:
            switch(y) {
                case 0:
                    return 3;
                case 1:
                    return 3;
                case 2:
                    return 2;
                case 3:
                    return 1;
                case 4:
                    return 2;
            }
        case 4:
            switch(y) {
                case 0:
                    return 2;
                case 1:
                    return 2;
                case 2:
                    return 3;
                case 3:
                    return 3;
                case 4:
                    return 1;
            }
    }
    return 0;
}
int main() {
    cin >> n >> na >> nb;
    for (int i = 1; i <= na; i++) cin >> a[i];
    for (int i = 1; i <= nb; i++) cin >> b[i];
    a[0] = a[na];
    b[0] = b[nb];
    for (int i = 1; i <= n; i++) {
        if (cmp(a[i % na], b[i % nb]) == 1) {
        } else if (cmp(a[i % na], b[i % nb]) == 2) {
            aa++;
        } else {
            ab++;
        }
    }
    cout << aa << " " << ab;
}