题目：
第一题
It is really amazing that the great historian Dr.K has recently found that about 10 Million years ago, in the area where is now called China, lived an ancient people. They may be considered as the first intelligent creature existed on the earth.

As Dr.K's investigation advance, he found unbelievably that they even developed some so-called mathematics during their evolvement. Dr.K spent his half life to understand this ancient people's counting system. Finally he got to know that:
1、They use only 7 digits to represent numbers, and numbers are as follow:
| -> 1
|| -> 2
||| -> 3
|||| -> 4
||||| -> 5
|||||| -> 6
||||||| -> 7
It is a pity that they have not developed "0" before these people disappeared.
2、If a number is greater than 7, they use digit place just as we do today. Between each two digit places, there is a ",".
eg:
|||||,|| -> 42 (5x8+2)
|||,||,|||| -> 212 (3*64+2*8+4)

In order to further his study, Dr.K wants to know what the sequences found from stones mean in today's counting system. He turns to you for help, and smart as you are, you should help this great historian,should not you?

Input

The first line of standard input contains an integer N, the number of test cases. N lines followed.
In each line a sequence is given, it is guaranteed that the length of the sequence will not exceed 1024 characters and the outcome number will not be greater than 1000000.

Output

For each case you are to output a line giving the number in today's decimal counting system.

Sample Input

3
|||||,||
|||,||,||||
||,|,|,|,||||

Sample Output

42
212
8780

为什么这段程序在i=１时是正确的，而在i>1时是错误的
#include <stdio.h>
#define MAX 54
void main()
{
char p;
int i,j,x;
float m=0,k=0;
char str[1024][MAX];
scanf("%d",&i);
printf("\n");
for(j=1;j<=i;j++)
{
scanf("%s",str[j]);
p=str[j][0];
while(p!='\0')
{
while(p!=',')
{k++;m++;
p=str[j][m];
if(p=='\0') goto loop;
}
m++;
k=k*8;
p=str[j][m];
}
loop: printf("%f\n",k);
}
}

如输入　１
　　　　　｜｜｜｜，｜｜｜｜
输出　３６　正确
而输入　２　　　　　　　输出
　　　　｜｜｜｜　　　　４
　　　　｜，｜｜　　　　５
　错误
令我很郁闷，哪位大侠帮帮忙啊！！！！！

谢谢!实在是昨天头脑短路了!!!!!!谢谢

#include<stdio.h>
#define MAX 54
void main()
{
int i,m,k,sum=0;
char str[MAX][80];
int num;
printf("please input num:");
scanf("%d",&num);
for(i=0;i<num;i++)
{
scanf("%s",str[i]);
m=0;
k=0;
sum=0;
for(k=0;str[i][k]!='\0';k++)
{
while(str[i][k]!=',' && str[i][k]!='\0')
{
m++;
k++;
}
sum=sum*8+m;
m=0;
}
printf("%d\n",sum);
}
}

我也写了这个程序，拿出来大家探讨一下：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define DECI 8
#define ARRAY_SIZE 4

char *ancient_num[ARRAY_SIZE]={NULL};
int input_num=0;
int moden_num[ARRAY_SIZE];

void input();
void calculate();
void output();
int d(int decimal);

void main()
{
input();
calculate();
output();
}

void input()
{
int i;

printf("How many ancient numbers you want to input?\n");
scanf("%d",&input_num);

if(input_num>ARRAY_SIZE||input_num==0)
{
printf("You can put %d ancient numbers!\n",ARRAY_SIZE);
exit(0);
}
else
for(i=1;i<=input_num;i++)
{
printf("%d :",i);
scanf("%s",ancient_num[i-1]=(char*)malloc(80));
putchar('\n');
}

}

void calculate()
{
int i,j=0;
int len,decimal;
int stick; /*记录每一位上的”|“*/
char *rev; /*把字符串倒转*/
for(i=0;i<input_num;i++)
{
rev=(char*)malloc(80);
strrev(ancient_num[i]);
len=strlen(ancient_num[i]);
decimal=0;
stick=0;
for(j=0;j<len+1;j++)
{
if(ancient_num[i][j]=='|')
{
stick++;
}
else if(ancient_num[i][j]==',')
{
moden_num[i]+=(d(decimal)*stick);
stick=0;
decimal++;
}
else if(j==len)
{
moden_num[i]+=d(decimal)*stick;
}
}
free(ancient_num[i]);

}

}

int d(int decimal)
{
int i,deci=1;
if(decimal==0)
return 1;
else
for(i=0;i<decimal;i++)
deci*=DECI;
return deci;
}

void output()
{
int i;
printf("anser:\n");
for(i=0;i<input_num;i++)
{
printf("%d \n",moden_num[i]);
}
}

又是英语