代码

// 1000i: traditional A+B, n=100 pairs.  No libc init, direct DuckInfo I/O.
// DuckInfo = 0x400000 + memory_hard_limit - 112 = 0x447FF90 for this problem.
typedef unsigned long long u64;
#define DUCKINFO 0x447FF90ULL
#define DI_IN   8
#define DI_INSZ 16
#define DI_OUT  24
#define DI_OUTSZ 40

static inline void rx(void) {
    register long rax __asm__("rax") = 60;
    register long rdi __asm__("rdi") = 0;
    __asm__ volatile("syscall" :: "a"(rax), "D"(rdi) : "rcx", "r11", "memory");
    __builtin_unreachable();
}

static inline long parse_long(const char **pp, const char *end) {
    const char *p = *pp;
    while (p < end && (*p == ' ' || *p == '\n' || *p == '\r' || *p == '\t')) p++;
    int neg = 0;
    if (p < end && *p == '-') { neg = 1; p++; }
    long v = 0;
    while (p < end && *p >= '0' && *p <= '9') { v = v * 10 + (*p - '0'); p++; }
    *pp = p;
    return neg ? -v : v;
}

static inline char *emit_ll(char *o, long v) {
    if (v < 0) { *o++ = '-'; v = -v; }
    char tmp[24]; int k = 0;
    do { tmp[k++] = char('0' + v % 10); v /= 10; } while (v);
    while (k) *o++ = tmp[--k];
    *o++ = '\n';
    return o;
}

extern "C" int __libc_start_main(int (*m)(int, char **, char **), int c, char **v,
                                 void (*i)(void), void (*f)(void), void (*l)(void)) {
    char *di = (char *)DUCKINFO;
    const char *in = *(const char **)(di + DI_IN);
    u64 insz = *(u64 *)(di + DI_INSZ);
    char *out = *(char **)(di + DI_OUT);
    const char *p = in, *end = in + insz;
    long n = parse_long(&p, end);
    char *o = out;
    for (long k = 0; k < n; k++) {
        long a = parse_long(&p, end);
        long b = parse_long(&p, end);
        o = emit_ll(o, a + b);
    }
    *(u64 *)(di + DI_OUTSZ) = (u64)(o - out);
    rx();
    return 0;
}
int main() { return 0; }

评测结果