洛谷 P1328 [NOIP2014 提高组] 生活大爆炸版石头剪刀布

题目大意

小A和小B，要进行 \(N\) 次猜拳，每次按照一定周期出拳，胜负情况如下：

求出小A和小B分别赢了几次。

思路

枚举 \(N\) 次猜拳，每次比较 \(a[powera]\) 与 \(b[powerb]\) （poewra 与 powerb 是 a 和 b 数组的索引，详见代码）。

CODE

#include<bits/stdc++.h>
using namespace std;
int a[2005], b[2005];
int main(){int n, na, nb, ansa  = 0, ansb = 0;cin >> n >> na >> nb;for (int i = 1; i <= na; i++) {cin >> a[i];}for (int i = 1; i <= nb; i++) {cin >> b[i];}int powera = 1, powerb = 1; for (int i = 1; i <= n; i++) {if (powera > na) { //超过周期长度的特判 powera = 1;}if (powerb > nb) {powerb = 1;}if (a[powera] == 0) {if (b[powerb] == 1) {ansb++;}if (b[powerb] == 2) {ansa++;}if (b[powerb] == 3) {ansa++;}if (b[powerb] == 4) {ansb++;}}if (a[powera] == 1) {if (b[powerb] == 0) {ansa++;}if (b[powerb] == 2) {ansb++;}if (b[powerb] == 3) {ansa++;}if (b[powerb] == 4) {ansb++;}}if (a[powera] == 2) {if (b[powerb] == 0) {ansb++;}if (b[powerb] == 1) {ansa++;}if (b[powerb] == 3) {ansb++;}if (b[powerb] == 4) {ansa++;}}if (a[powera] == 3) {if (b[powerb] == 0) {ansb++;}if (b[powerb] == 1) {ansb++;}if (b[powerb] == 2) {ansa++;}if (b[powerb] == 4) {ansa++;}}if (a[powera] == 4) {if (b[powerb] == 0) {ansa++;}if (b[powerb] == 1) {ansa++;}if (b[powerb] == 2) {ansb++;}if (b[powerb] == 3) {ansb++;}}powera++;powerb++;} cout << ansa << " " << ansb << endl;return 0;
}