Google Code Jam 2010 Qualification Round 2012 Problem B && C

B题

当p=0时所有的总分都在sum1++，

当p=1时，总分>=1时在sum1++，总分==0，是无论如何都满足不了题意（坑点）；

总分>=29时是没有surprising情况的，无论p取何值都在sum1++；

#include<cstdio>
#include<cstring>
using namespace std;
int main(){
int n,i,j,T,t,s,p,a;
char ch;
freopen( "B-large.in", "r", stdin );
freopen( "output1.out", "w", stdout );
scanf("%d",&T);
for(t=1;t<=T;t++){
printf("Case #%d: ",t);
int sum1=0,sum2=0,sum=0;
scanf("%d %d %d",&n,&s,&p);
for(i=1;i<=n;i++){
scanf("%d",&a);
if(a>=p*3-2)
sum1++;
else if(a==0 && p==1)
;
else if(a>=p*3-4)
sum2++;
}
if(sum2>s)
printf("%d\n",s+sum1);
else
printf("%d\n",sum1+sum2);

}
}

C题

这个题有个易错点。就是1212这种自身带循环节的情况。

易分析如果一个数移位后得到自身，那么此数一定有循环节。

#include<cstdio>
#include<cstring>
#include<math.h>
using namespace std;
int main(){
int i,j,T,t,a,b,n,p;
char ch;
freopen("C-large.in","r",stdin);
freopen("output1.out","w",stdout);
scanf("%d",&T);
for(t=1;t<=T;t++){
printf("Case #%d: ",t);
scanf("%d %d",&a,&b);
if(a<10)
printf("0\n");
else{
int n=(int)log10((double)b)+1;
int con=(int )pow((double)10,(double)n);
int sum=0;
for(i=a;i<=b;i++){
int tem1,tem2=a;
p=10;
while(1){
tem1=i%p;
tem2=i/p;
if(tem1*(con/p)+tem2==i)     //////
break;
if(tem1*(con/p)+tem2>i && tem1*(con/p)+tem2<=b)
sum++;
p*=10;
}
}
printf("%d\n",sum);
}
}
}