快速读入

int read() {
    int f=1,x=0;char ch=getchar();
    while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
    return f*x;
}

加入下面的速度更快：

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;

cin 关闭流同步

关闭同步后不能用 scanf 和 printf ！

ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);

关于 cout 输出小数点：

#include <iomanip>
int main() {
    const double value = 12.3456789;
    cout << value << endl; // 默认以6精度，所以输出为 12.3457
    cout << setprecision(4) << value << endl;
    // 改成4精度，所以输出为12.35

    cout << setprecision(8) << value << endl;
    // 改成8精度，所以输出为12.345679

    cout << fixed << setprecision(4);
    cout << value << endl;
    // fixed：固定点方式显示，精度指的是小数位，输出为12.3457

    cout << value << endl;
    // fixed和setprecision的作用还在，依然显示12.3457

    cout.unsetf( ios::fixed ); cout << value << endl;
    // 去掉fixed，精度恢复成整个数值的有效位数，显示为12.35

    cout.precision( 6 ); cout << value << endl;
    // 恢复成原来的样子，输出为12.3457
}

__int128 的输入输出重载

istream& operator >> (istream& stream, __int128 &v) {
    string s; cin >> s; int l = (int)s.length(), i = 0, f = 1;
    for (; i < l; i++) {
        if (s[i] == '-') f = -1;
        if (s[i] >= '0' && s[i] <= '9') break;
    }
    for (v = 0; i < l; i++) v = (v << 3) + (v << 1) + (s[i] ^ 48);
    return stream;
}

ostream& operator << (ostream& stream, __int128 &v) {
    __int128 x = v;
    if (v < 0) {stream << '-'; x = -x;}
    vector<int> t;
    if (v == 0) stream << "0";
    while (x) {t.push_back(x % 10); x /= 10;}
    for (auto it = t.rbegin(); it != t.rend(); it++) stream << *it;
    return stream;
}

大整数取模

读入时取模

int getmod(int p) {
    int b = 0; char c;
    while (!isdigit(c = getchar()));
    for (; isdigit(c); c = getchar()) {
        b = b * 10 + c - '0';
        if (b >= p) b %= p;
    }
    return b;
}

读入后取模

ll Int(ll p, char *ch) {
    ll res = 0;
    int pos = 0;
    while (!isdigit(ch[pos])) pos++;
    while (isdigit(ch[pos])) {
       res = (res * 10 + ch[pos] - '0') % p;
        pos++;
    }
    return res;
}