C.445 - Marvelous Mazes
2013-06-17 00:00
716 查看
此题最重要的是怎样计算打印个数 .
if(str[i]>='0'&&str[i]<='9') sum+=str[i]-'0'; 这段代码即for循环的综合运用让人恍然大悟,确实不错,为我解决了很多问题.
11X21b1X
=>2X3b1X
int main()
{
char str[133];
while(gets(str)!=NULL)
{
int len=strlen(str);
int sum=0;
for(int i=0; i<len; i++)
{
if(str[i]>='0'&&str[i]<='9') sum+=str[i]-'0';
else if(str[i]>='A'&&str[i]<='Z')
{
for(int j=0; j<sum; j++)
putchar(str[i]);
sum=0;
}
else if(str[i]=='b')
{
for(int j=0; j<sum; j++)
putchar(' ');
sum=0;
}
else if(str[i]=='*')
{
for(int j=0; j<sum; j++)
putchar('*');
sum=0;
}
else if(str[i]=='!') putchar('\n');
}
putchar('\n');
}
return 0;
}
if(str[i]>='0'&&str[i]<='9') sum+=str[i]-'0'; 这段代码即for循环的综合运用让人恍然大悟,确实不错,为我解决了很多问题.
11X21b1X
=>2X3b1X
int main()
{
char str[133];
while(gets(str)!=NULL)
{
int len=strlen(str);
int sum=0;
for(int i=0; i<len; i++)
{
if(str[i]>='0'&&str[i]<='9') sum+=str[i]-'0';
else if(str[i]>='A'&&str[i]<='Z')
{
for(int j=0; j<sum; j++)
putchar(str[i]);
sum=0;
}
else if(str[i]=='b')
{
for(int j=0; j<sum; j++)
putchar(' ');
sum=0;
}
else if(str[i]=='*')
{
for(int j=0; j<sum; j++)
putchar('*');
sum=0;
}
else if(str[i]=='!') putchar('\n');
}
putchar('\n');
}
return 0;
}
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