Hard-decision decoding scheme
Codeword length (n) : 31 symbols.
Message length (k) : 19 symbols.
Error correction capability (t) : 6 symbols
One symbol represents 5 bit.
Uses GF(2^5) with primitive polynomial p(x) = X^5 X^2 + 1
Generator polynomial, g(x) = a^15 a^21*X + a^6*X^2 + a^15*X^3 + a^25*X^4 + a^17*X^5 + a^18*X^6 + a^30*X^7 + a^20*X^8 + a^23*X^9 + a^27*X^10 + a^24*X^11 + X^12. Note: a = alpha, primitive element in GF(2^5) and a^i is root of g(x) for i = 19, 20, ..., 30.
Uses Verilog description with synthesizable RTL modelling.
Consists of 5 main blocks: SC (Syndrome Computation), KES (Key Equation Solver), CSEE (Chien Search and Error Evaluator), Controller and FIFO Register.
標簽：
symbols
length
Hard-decision
Codeword
上傳時間：
2014-07-08
上傳用戶：曹云鵬
動態規劃的方程大家都知道，就是
f[i,j]=min{f[i-1,j-1],f[i-1,j],f[i,j-1],f[i,j+1]}+a[i,j]
但是很多人會懷疑這道題的后效性而放棄動規做法。
本來我還想做Dijkstra，后來變了沒二十行pascal就告訴我數組越界了……（dist:array[1..1000*1001
div 2]...）
無奈之余看了xj_kidb1的題解，剛開始還覺得有問題，后來豁然開朗……
反復動規。上山容易下山難，我們可以從上往下走，最后輸出f[n][1]。
xj_kidb1的一個技巧很重要，每次令f[i][0]=f[i][i],f[i][i+1]=f[i][1]（xj_kidb1的題解還寫錯了）
標簽：
動態規劃
方程
家
上傳時間：
2014-07-16
上傳用戶：libinxny
1．Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and
another integer x, determines whether or not there exist two elements in S whose sum is exactly x. (Implement exercise 2.3-7.)
#include<stdio.h>
#include<stdlib.h>
void merge(int arr[],int low,int mid,int high){
int i,k;
int *tmp=(int*)malloc((high-low+1)*sizeof(int));
int left_low=low;
int left_high=mid;
int right_low=mid+1;
int right_high=high;
for(k=0;left_low<=left_high&&right_low<=right_high;k++)
{
if(arr[left_low]<=arr[right_low]){
tmp[k]=arr[left_low++];
}
else{
tmp[k]=arr[right_low++];
}
}
if(left_low<=left_high){
for(i=left_low;i<=left_high;i++){
tmp[k++]=arr[i];
}
}
if(right_low<=right_high){
for(i=right_low;i<=right_high;i++)
tmp[k++]=arr[i];
}
for(i=0;i<high-low+1;i++)
arr[low+i]=tmp[i];
}
void merge_sort(int a[],int p,int r){
int q;
if(p<r){
q=(p+r)/2;
merge_sort(a,p,q);
merge_sort(a,q+1,r);
merge(a,p,q,r);
}
}
int main(){
int a[8]={3,5,8,6,4,1,1};
int i,j;
int x=10;
merge_sort(a,0,6);
printf("after Merging-Sort:\n");
for(i=0;i<7;i++){
printf("%d",a[i]);
}
printf("\n");
i=0;j=6;
do{
if(a[i]+a[j]==x){
printf("exist");
break;
}
if(a[i]+a[j]>x)
j--;
if(a[i]+a[j]<x)
i++;
}while(i<=j);
if(i>j)
printf("not exist");
system("pause");
return 0;
}
標簽：
c語言
算法
排序
上傳時間：
2017-04-01
上傳用戶：糖兒水嘻嘻
