【BZOJ 2456】mode

Qizy 其他, 杂项, 解题报告

相关链接

题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2456
神犇题解：https://oi.men.ci/bzoj-2456/

解题报告

考虑将不同的数抵消掉
因为题目所求的数多于一半
所以剩下的就一定是答案啦！

Code

#include<cstdio>
#define LL long long

int main() {
	int n,tot=0; LL num,cur;
	for (scanf("%d",&n);n;n--) {
		scanf("%lld",&cur);
		if (num != cur) {
			if (tot) tot--;
			else tot = 1, num = cur;
		} else tot++;
	} printf("%lld\n",num);
	return 0;
}

—– UPD 2016.12.29 —–
这题似乎还有一个做法：

考虑将原数分解成二进制
统计每一位二进制上0多还是1多

酱紫的话，最后在在二进制下逐位确定答案就好啦！

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