Sicily 1920 Divide The Stones

每次最少分出一个，则最后每个堆的 石头数量都是1， 共有sum(a[i])堆。而分出这些石头共需要sum(a[i] - 1)次操作。对这个操作数进行奇偶判断，若为奇数则Alice胜，否则Bob必胜。

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>using namespace std;int main(){int t, n;scanf("%d", &t);while (t--){int sum = 0, temp;scanf("%d", &n);for (int i = 0; i < n; i++){scanf("%d", &temp);sum += (temp - 1);}if (sum & 1)printf("Alice\n");elseprintf("Bob\n");}}