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/*
*题目链接:https://www.nowcoder.com/acm/contest/186/A
*分析:奇数x分为(1,x - 1)时为最优解,偶数等待合并
*若奇数x == 1,忽略此数
*a个奇数分解花费a次,b个偶数合并花费(b - 1)次
*a个奇数分解后有a + b个偶数,b是原始的偶数个数
*综上所述,总的可执行过程为a +(a + b - 1)= 2 * a + b - 1次
*注意,若2 * a + b == 0,我们要输出Bob
**/
#include <stdio.h>
int main() {
int n;
while (scanf("%d", &n) != EOF) {
int a = 0, b = 0;
for (int i = 0; i < n; ++i) {
int x;
scanf("%d", &x);
if ((x & 1) && x > 2) a++;
else if (x % 2 == 0) b++;
}
int num = (2 * a + b) - 1;
if (!a && !b) num = 0;
if (num % 2) puts("Alice");
else puts("Bob");
}
return 0;
}