代码搜索结果
找到约 10,000 项符合
Input 的代码
readme
To do a visualization type:
ch3-04.out <
input1:
x = 3 * 2
y = x + 1
3 * y
input2:
3 * (2 + 4)
x = 24
y = x + 1
3 * y
2 + x + y
input3:
foo = 12 * 2
foo / 5
thisisareallylongvar
adcint.map.eqn
-- Copyright (C) 1991-2005 Altera Corporation
-- Your use of Altera Corporation's design tools, logic functions
-- and other software and tools, and its AMPP partner logic
-- functions, and any o
binin.so
[
.B \-if|\-input-file|< " name"
]
[
.B \-input-byteorder|-input-endian " big|little"
]
insert_wm.m
function [input_string,out] = insert_WM(input,blk,alpha,WM)
% This function will insert the watermark int he input
% audio stream. The input variable are
% input = input audio stream
% block = in
in_file.c
#include "common.h"
#include "input.h"
FILE *input_fd=NULL;
extern int input_size;
extern int input_remain;
extern int ring_pos;
extern unsigned char ring_buf[RING_BUF_SIZE];
int input_open(char *f
rfc336.txt
NIC 9929 Ira Cotton
NWG/RFC #336 Mitre
Level 0 Graphic Input Protocol 5 May 197
4.6readme.txt
要点：
将n个整数排成一排，进行如下操作，每次删除两个数，然后在数列中加入一个数a*b+1，如此下去直到剩下一个数，求所得到的数的最大值和最小值的差值。通过有效的利用空间完成了空间的合理分配。
思路
用贪心的思想，假设有三个数a，a+k1，a+k1+k2，k1>0,k2>0;
有已下几种组合：
(1)(a*b+1)*c+1=(a*(a+k1)+1)*(a+k1+k2)+1=a*a*a+2 ...
rfc336.txt
NIC 9929 Ira Cotton
NWG/RFC #336 Mitre
Level 0 Graphic Input Protocol 5 May 197
add.map.eqn
-- Copyright (C) 1991-2005 Altera Corporation
-- Your use of Altera Corporation's design tools, logic functions
-- and other software and tools, and its AMPP partner logic
-- functions, and any o
test1.map.eqn
-- Copyright (C) 1991-2005 Altera Corporation
-- Your use of Altera Corporation's design tools, logic functions
-- and other software and tools, and its AMPP partner logic
-- functions, and any o