Problem F
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 91 Accepted Submission(s) : 17
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Problem Description
We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example, S(893)=3, S(114514)=6.
You want to make a consecutive integer sequence starting from number m (m,m+1,...). But you need to pay S(n)·k to add the number n to the sequence.
You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.
You want to make a consecutive integer sequence starting from number m (m,m+1,...). But you need to pay S(n)·k to add the number n to the sequence.
You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.
Input
The first line contains three integers w (1≤w≤10^16), m (1≤m≤10^16), k (1≤k≤10^9).
Output
The first line should contain a single integer — the answer to the problem.
Sample Input
9 1 1 77 7 7 114 5 14 1 1 2
Sample Output
9 7 6 0
————————————————————————————————————————————————————————————————————
（ps：题真恶心啊。。。写了两发暴力都超时）
1.a=w/k简化运算
2.
代码如下：
AC代码：
#include<stdio.h>
int cal(long long n)
{
int count = 0;
while(n)
{
n /= 10;
count++;
}
return count;
}
int main()
{
long long ten;
char a[20];
long long k,count,num,l;
long long w,m,i,j;
long long sum;
while(~scanf("%I64d%I64d%I64d",&w,&m,&k))
{
ten = 1;
sum = 0;
count = 0;
j = w/k;
num = cal(m);
for(i = 0; i < num; i++)
{
ten *= 10;
}
while(1)
{
if((ten - m)*num < j)
{
j -= (ten - m)*num;
count += (ten - m);
}
else
{
count += (j/num);
break;
}
m = ten;
ten *= 10;
num++;
}
printf("%I64d\n",count);
}
return 0;
}
暴力超时：
#include<stdio.h>
#include<string.h>
#include<math.h>
int fun(long long n)
{
int v=10,t=0;
while(n){
n=n/10;
t++;
}
return t;
}
int main()
{
long long w,m,s,k,i,a,x,q,t,count;
int v;
while(~scanf("%lld %lld %lld",&w,&m,&k))
{
a=w/k;
t=0;s=0;count=0;
if(m<=9)
while(m)
{
m=m%10;
t++;
if(m&&(s+=(10-m)*t<=a)) s+=(10-m)*t,count++;
}
/* s=0;
for(i=m;;i++)
{
s+=fun(i)*k;
if(s>w) break;
}
*/
printf("%d\n",count);
}
return 0;
}
ps：注释是第一次超时，剩下的是第二次超时。。。
总体思路就是暴力直接过。还是太年轻吧。。。
人一我百，人十我万！！！