感觉三进制不进位加法应该能解决这个问题

用暴力法实现了一下，不知道怎么优化

更新：发现负数处理问题，调试发现直接(unsigned long long)相当于(unsigned long long)(long long)，会扩展符号位，要(unsigned long long)(unsigned)才会用零填充高位，坑了。。。已修正

<code class="language-cpp">#include <stdio.h>

int main(int argc, char *argv[])
{
	int count, nums[256];
	scanf("%d", &count);
	for (int i = 0; i < count; i++)
		scanf("%d", &nums[i]);

	unsigned long long prevresult = count > 0 ? (unsigned)nums[0] : 0;
	for (int i = 1; i < count; i++)
	{
		unsigned long long prev = 0, result = 0;
		for (unsigned long long j = 3; j <= 10460353203ull; j *="3)" { unsigned long now="(unsigned)nums[i]" % + prevresult j; result - prev) prev="now;" } printf("%d\n", (int)prevresult); return 0; < code></=></stdio.h></code>