Instead of finding the longest common
subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽：
the
subsequence
determine
Instead
上傳時間：
2013-12-17
上傳用戶：evil
#include<reg51.h>
#define uchar unsigned char
#define uint unsigned int
uint i,j;
sbit dula=P2^6;
sbit wela=P2^7;
uchar code table[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,
0x7d,0x07,0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
void main()
{
j=0;
i=0;
TMOD=0X01;
TH0=(65536-50000)/256;
TL0=(65536-50000)%6;
EA=1;
ET0=1;
TR0=1;
while(1);
}
void time0() interrupt 1
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%6;
i++;
if(i==15)
{
P0=table[j];
dula=1;
dula=0;
P0=0XC0;
wela=1;
wela=0;
j++;
i=0;
if(j==16)
{
j=0;
}
}
}
標簽：
用定時器以間隔500MS在6位數碼管上依次顯示0、1、 2、3….C、D、E、F，重復。
上傳時間：
2016-02-11
上傳用戶：嬌縱Pamper
#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define NULL 0
#define MaxSize 30
typedef struct athletestruct /*運動員*/
{
char name[20];
int score; /*分數*/
int range; /**/
int item; /*項目*/
}ATH;
typedef struct schoolstruct /*學校*/
{
int count; /*編號*/
int serial; /**/
int menscore; /*男選手分數*/
int womenscore; /*女選手分數*/
int totalscore; /*總分*/
ATH athlete[MaxSize]; /**/
struct schoolstruct *next;
}SCH;
int nsc,msp,wsp;
int ntsp;
int i,j;
int overgame;
int serial,range;
int n;
SCH *head,*pfirst,*psecond;
int *phead=NULL,*pafirst=NULL,*pasecond=NULL;
void create();
void input ()
{
char answer;
head = (SCH *)malloc(sizeof(SCH)); /**/
head->next = NULL;
pfirst = head;
answer = 'y';
while ( answer == 'y' )
{
Is_Game_DoMain:
printf("\nGET Top 5 when odd\nGET Top 3 when even");
printf("\n輸入運動項目序號 (x<=%d):",ntsp);
scanf("%d",pafirst);
overgame = *pafirst;
if ( pafirst != phead )
{
for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ )
{
if ( overgame == *pasecond )
{
printf("\n這個項目已經存在請選擇其他的數字\n");
goto Is_Game_DoMain;
}
}
}
pafirst = pafirst + 1;
if ( overgame > ntsp )
{
printf("\n項目不存在");
printf("\n請重新輸入");
goto Is_Game_DoMain;
}
switch ( overgame%2 )
{
case 0: n = 3;break;
case 1: n = 5;break;
}
for ( i = 1 ; i <= n ; i++ )
{
Is_Serial_DoMain:
printf("\n輸入序號 of the NO.%d (0<x<=%d): ",i,nsc);
scanf("%d",&serial);
if ( serial > nsc )
{
printf("\n超過學校數目,請重新輸入");
goto Is_Serial_DoMain;
}
if ( head->next == NULL )
{
create();
}
psecond = head->next ;
while ( psecond != NULL )
{
if ( psecond->serial == serial )
{
pfirst = psecond;
pfirst->count = pfirst->count + 1;
goto Store_Data;
}
else
{
psecond = psecond->next;
}
}
create();
Store_Data:
pfirst->athlete[pfirst->count].item = overgame;
pfirst->athlete[pfirst->count].range = i;
pfirst->serial = serial;
printf("Input name:) : ");
scanf("%s",pfirst->athlete[pfirst->count].name);
}
printf("\n繼續輸入運動項目(y&n)？");
answer = getchar();
printf("\n");
}
}
void calculate() /**/
{
pfirst = head->next;
while ( pfirst->next != NULL )
{
for (i=1;i<=pfirst->count;i++)
{
if ( pfirst->athlete[i].item % 2 == 0 )
{
switch (pfirst->athlete[i].range)
{
case 1:pfirst->athlete[i].score = 5;break;
case 2:pfirst->athlete[i].score = 3;break;
case 3:pfirst->athlete[i].score = 2;break;
}
}
else
{
switch (pfirst->athlete[i].range)
{
case 1:pfirst->athlete[i].score = 7;break;
case 2:pfirst->athlete[i].score = 5;break;
case 3:pfirst->athlete[i].score = 3;break;
case 4:pfirst->athlete[i].score = 2;break;
case 5:pfirst->athlete[i].score = 1;break;
}
}
if ( pfirst->athlete[i].item <=msp )
{
pfirst->menscore = pfirst->menscore + pfirst->athlete[i].score;
}
else
{
pfirst->womenscore = pfirst->womenscore + pfirst->athlete[i].score;
}
}
pfirst->totalscore = pfirst->menscore + pfirst->womenscore;
pfirst = pfirst->next;
}
}
void output()
{
pfirst = head->next;
psecond = head->next;
while ( pfirst->next != NULL )
{
// clrscr();
printf("\n第%d號學校的結果成績:",pfirst->serial);
printf("\n\n項目的數目\t學校的名字\t分數");
for (i=1;i<=ntsp;i++)
{
for (j=1;j<=pfirst->count;j++)
{
if ( pfirst->athlete[j].item == i )
{
printf("\n %d\t\t\t\t\t\t%s\n %d",i,pfirst->athlete[j].name,pfirst->athlete[j].score);break;
}
}
}
printf("\n\n\n\t\t\t\t\t\t按任意建 進入下一頁");
getchar();
pfirst = pfirst->next;
}
// clrscr();
printf("\n運動會結果:\n\n學校編號\t男運動員成績\t女運動員成績\t總分");
pfirst = head->next;
while ( pfirst->next != NULL )
{
printf("\n %d\t\t %d\t\t %d\t\t %d",pfirst->serial,pfirst->menscore,pfirst->womenscore,pfirst->totalscore);
pfirst = pfirst->next;
}
printf("\n\n\n\t\t\t\t\t\t\t按任意建結束");
getchar();
}
void create()
{
pfirst = (struct schoolstruct *)malloc(sizeof(struct schoolstruct));
pfirst->next = head->next ;
head->next = pfirst ;
pfirst->count = 1;
pfirst->menscore = 0;
pfirst->womenscore = 0;
pfirst->totalscore = 0;
}
void Save()
{FILE *fp;
if((fp = fopen("school.dat","wb"))==NULL)
{printf("can't open school.dat\n");
fclose(fp);
return;
}
fwrite(pfirst,sizeof(SCH),10,fp);
fclose(fp);
printf("文件已經成功保存\n");
}
void main()
{
system("cls");
printf("\n\t\t\t 運動會分數統計\n");
printf("輸入學校數目 (x>= 5):");
scanf("%d",&nsc);
printf("輸入男選手的項目(x<=20):");
scanf("%d",&msp);
printf("輸入女選手項目(<=20):");
scanf("%d",&wsp);
ntsp = msp + wsp;
phead = (int *)calloc(ntsp,sizeof(int));
pafirst = phead;
pasecond = phead;
input();
calculate();
output();
Save();
}
標簽：
源代碼
上傳時間：
2016-12-28
上傳用戶：150501
