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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 numbern 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 ≤ 1016), m (1 ≤ m ≤ 1016), k (1 ≤ k ≤ 109).
Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Output
The first line should contain a single integer — the answer to the problem.
Sample Input
9 1 1
9
77 7 7
7
114 5 14
6
1 1 2
0
1 #include <stdio.h> 2 #include <string.h> 3 long long SS(long long a) 4 { 5 long long x=0; 6 while(a>0) 7 { 8 a=a/10; 9 x++; 10 //printf("%I64d\n",a); 11 } 12 return x; 13 } 14 15 long long num(long long x) 16 { 17 long long i,y=1; 18 for(i=1;i<=x;i++) 19 y=y*10; 20 return y; 21 } 22 int main() 23 { 24 long long w,m,k; 25 int i,j; 26 while(scanf("%I64d %I64d %I64d",&w,&m,&k)!=EOF) 27 { 28 long long n=0,nu,ww,mm; 29 long long oo=SS(m); 30 nu=num(oo); 31 ww=w/k; 32 while(ww>0) 33 { 34 mm=ww/oo; 35 n=n+mm; 36 if(n>=nu-m) 37 { 38 mm=mm-(n-nu+m); 39 n=nu-m; 40 ww=ww-oo*mm; 41 oo++; 42 nu=num(oo); 43 } 44 else 45 { 46 ww=ww-oo*mm; 47 } 48 if(mm==0) 49 break; 50 //printf("%I64d %I64d %I64d",ww,); 51 } 52 printf("%I64d\n",n); 53 } 54 return 0; 55 }
CodeForces 373B Making Sequences is Fun
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原文地址：http://www.cnblogs.com/cyd308/p/4771495.html
