关于快读

普通的快读大家应该都会写，我就不再赘述，这里讲一下用fread的快读。

fread

首先函数原型

size_t fread(void *ptr, size_t size, size_t nmemb, FILE *stream);

函数fread从stream流中读取nmemb个长度为size字节大小的数据，并将它们存储在ptr给定的位置，返回读取的字节数。

比如fread(buf, 1, 10, stdin);就是从标准输入流中读取10个1字节大小的字符，存在buf数组里。

因为它是一下子读一大堆，所以比较快，那么怎么使用呢？

使用

首先定义一个字符数组，大小为\(2^{20}\) (1 mb), 作为缓存，每次读取1 mb数据存在数组里，读数字的时候扫描数组，像普通快读一样转成数字，读完缓存就再读 1 mb。

char buf[1<<20], *p1, *p2; //buf 缓存， p1 为指向缓存开头的指针， p2 为指向缓存结束的指针
char gc() { 
    if (p1 == p2) { // 判断是否第一次执行或读取完缓存
        p1 = buf; // 重置开头指针
        p2 = buf + fread(buf, 1, 1<<20, stdin); //读取数据， 重置结尾指针
        if (p1 == p2) return EOF;
    }
    return *p1++; // 返回第一个字符并指针后移
}
inline int read() { //普通的快读
    int f = 1, x = 0;
    char c = gc();
	while (c < '0' || c > '9') {
	    if (c == '-') f = -1;
	    c = gc();
	}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc();
	return f * x;
}

去注释，压行版本

char buf[1<<20], *p1, *p2;
char gc() { return p1 == p2 ? p2 = buf + fread(p1 = buf, 1, 1<<20, stdin), (p1 == p2) ? EOF : *p1++ : *p1++; }
inline int read(int f = 1, char c = gc(), int x = 0) {
	while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = gc();
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc();
	return f * x;
}

宏函数版本

char buf[1<<20],*p1,*p2;
#define gc() (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? EOF : *p1++) : *p1++)
#define read() ({\
    int x = 0, f = 1;\
    char c = gc();\
    while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = gc();\
    while(c >= '0' && c <= '9') x = x * 10 + (c & 15), c = gc();\
    f * x;\
})

宏函数一定会内联，比如以下代码

#include <cstdio>
char buf[1<<20],*p1,*p2;
#define gc() (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? EOF : *p1++) : *p1++)
#define read() ({\
    int x = 0, f = 1;\
    char c = gc();\
    while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = gc();\
    while(c >= '0' && c <= '9') x = x * 10 + (c & 15), c = gc();\
    f * x;\
})
int a[100];
int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; i++) a[i] = read();
}

就会展开为

#include <cstdio>
char buf[1<<20],*p1,*p2;
#define gc() (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? EOF : *p1++) : *p1++)
#define read() ({\
    int x = 0, f = 1;\
    char c = gc();\
    while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = gc();\
    while(c >= '0' && c <= '9') x = x * 10 + (c & 15), c = gc();\
    f * x;\
})
int a[100];
int main() {
    int n = ({ int x = 0, f = 1; char c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); while(c >= '0' && c <= '9') x = x * 10 + (c & 15), c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); f * x;}), m = ({ int x = 0, f = 1; char c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); while(c >= '0' && c <= '9') x = x * 10 + (c & 15), c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); f * x;});
    for (int i = 1; i <= n; i++) a[i] = ({ int x = 0, f = 1; char c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); while(c < '0' || c > '9') f = (c == '-') ? -1 : 1, c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); while(c >= '0' && c <= '9') x = x * 10 + (c & 15), c = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 1 << 20, stdin), p1 == p2 ? (-1) : *p1++) : *p1++); f * x;});
}