Input and output hang test template

Foreword

Playing newcode reproduce and Shanghai race when the two encountered a very similar time-out situation, in particular the proper way the question suggests special "Please use a more efficient input mode", misled me for a long time ,,,,
well-known standardized input and output ( scanf, printf) faster than the input and output stream (cin, cout), the reason can be found in a lot of Bowen, where the election a
https://www.cnblogs.com/xienaoban/p/6798095.html

Faster is linked to the input and output, because getchar () faster than Scanf,
so with the following formula:
input speed: getchar> scanf> cin

To test

First, randomly generated one million integers, write to the specified file

#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    srand(time(NULL));
    freopen("00out.txt", "w", stdout);
    int n;
    int num = 1000000;
    while(num--)
    {
        n = rand() % 1000;
        cout << n << ' ';
    }
    return 0;
}

Then, each linked to the input (getchar), scanf, cin read one million integer numbers, see the runtime

#include <ctime>
#include <cstdio>
#include <iostream>
using namespace std;
const int N = 1000005;
int a[N];
int b[N];

int Scan()     ///输入外挂
{
    int res = 0, ch ,flag = 0;
    if( (ch = getchar()) == '-' )
        flag = 1;
    else if(ch >= '0' && ch <= '9')
        res = ch - '0';
    while((ch = getchar()) >= '0' && ch <= '9')
        res = res * 10 + ch - '0';
    return flag ? -res : res;
}
void Out(int a)    ///输出外挂
{
    if(a < 0)
    {
        putchar('-');
        a *= -1;
    }
    if(a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

int main()
{
    clock_t start, over;
    double  duration;

    freopen("00out.txt", "r", stdin);
    start = clock();
    for(int i = 0; i < 1000000; ++i)
        a[i] = Scan();
    over = clock();
    duration = double(over - start) / CLOCKS_PER_SEC;
    cout << duration << " (s)" << '\n';
    fclose(stdin);

    freopen("00out.txt", "r", stdin);
    start = clock();
    for(int i = 0; i < 1000000; ++i)
        scanf("%d", &a[i]);
    over = clock();
    duration = double(over - start) / CLOCKS_PER_SEC;
    cout << duration << " (s)" << '\n';
    fclose(stdin);

    freopen("00out.txt", "r", stdin);
    start = clock();
    for(int i = 0; i < 1000000; ++i)
        cin >> a[i];
    over = clock();
    duration = double(over - start) / CLOCKS_PER_SEC;
    cout << duration << " (s)" << '\n';
    fclose(stdin);
}

Then

Ah ,,, so enter quickly hung (but rarely used, with this card really happened much the past)

O hang template

int Scan()     ///输入外挂
{
    int res = 0, ch ,flag = 0;
    if( (ch = getchar()) == '-' )
        flag = 1;
    else if(ch >= '0' && ch <= '9')
        res = ch - '0';
    while((ch = getchar()) >= '0' && ch <= '9')
        res = res * 10 + ch - '0';
    return flag ? -res : res;
}
void Out(int a)    ///输出外挂
{
    if(a < 0)
    {
        putchar('-');
        a *= -1;
    }
    if(a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

Read into the double hung on (but less than)

#include <cstdio>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;

double Scan()
{
    int ch, pos = 1, num = 0;///num为字符串总长度，pos为小数点排第几位
    double res = 0;
    bool flag = 0, found = 0;

    if( (ch = getchar()) == '-' )
        flag = 1;
    else if(ch >= '0' && ch <= '9')
        res = ch - '0';
    num++;

    while((ch = getchar()) >= '0' && ch <= '9' || ch == '.')
        {
            num++;
            if(ch == '.')
            {
                found = 1;
                pos++;
            }
            else
            {
                res = res * 10 + ch - '0';
                if(!found)
                    pos++;
            }
        }
    res *= pow(0.1, num - pos);
    return flag ? -res  : res ;
}

int main()
{
    double ans;
    while(ans = Scan())
    {
        cout << fixed << std::setprecision(10) << ans << '\n';
    }
    return 0;
}

Precautions

scanf("%d", &n);
getchar();
for(int i = 0; i < n; ++i)
    a[i] = Scan();

Getchar middle of the surface did not affect the time () may not be combined with the output of the test, but do question will give you off WA