代码

unsigned solve(int n, char *s) {
	unsigned int A[1000000] = {};
	unsigned int p = 1000001;
	A[500000] = 1;
	int i = 0;
	while (1) {
		switch (s[i++]) {
		case 0: goto end;
		case '(': --p; break;
		case ')': A[++p/2-1] = 0; break;
		case '?':
			unsigned* B = --p/2 + A;
			int m = 266670 - i;
			if (m>i) m=i;
			m /= 2;
			for (int j=0; j<m; j+=16) {
				asm volatile(
					"vmovupd 4(%0), %%ymm0; vmovupd 36(%0), %%ymm1;"
					"vmovupd 0(%0), %%ymm2; vmovupd 32(%0), %%ymm3;"
					"vpaddd %%ymm2, %%ymm0, %%ymm2; vpaddd %%ymm3, %%ymm1, %%ymm3;"
					"vmovupd %%ymm2, 0(%0); vmovupd %%ymm3, 32(%0);"
					: : "r"(B+j): "memory");
			}
	}	//printf("%d %d | %d %d %d %d\n", A[p/2-2], A[p/2-1], A[p/2], A[p/2+1], A[p/2+2], A[p/2+3]);
} end:;
	return A[p/2] & ~(1<<19);
}

评测结果