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my solution似乎不太难呢
好像有个叫做O(n)扫描的东西
#include <stdio.h>
long n;
void work()
{
char s[2000];
long l,i;
long ans;
scanf("%s",s);
l=strlen(s);
s[l]=10;
ans=0;
for(i=0;i<l;i++)
{
ans++;
if(s[i]!=s[i+1])
{
printf("%ld%ld",ans,s[i]-'0');
ans=0;
}
}
printf("\n");
}
int main()
{
long i;
scanf("%ld",&n);
for(i=1;i<=n;i++)
work();
return 0;
}
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