Codeforces Round #219 (Div. 2) B. Making Sequences is Fun
2 seconds
256 megabytes
standard input
standard output
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 payS(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.
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, coutstreams or the %I64d specifier.
The first line should contain a single integer — the answer to the problem.
9 1 1
9
77 7 7
7
114 5 14
6
1 1 2
0
多么典型的二分枚举呀,不过需要注意小细节。
#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef unsigned long long LL ; LL dp[] ;
LL num[] ; void init(){
num[] = ;
for(int i = ; i <= ; i++)
num[i] = num[i - ] * ;
for(int i = ; i <= ; i++)
dp[i] = i * (num[i+] - num[i]) ;
} LL get_all_S(LL x){
LL sum = ;
int i ;
for(i = ; i <= ; i++){
if(x >= num[i])
sum += dp[i-] ;
else
break ;
}
sum += (i-)*(x-num[i-]+) ;
return sum ;
} LL M ,W , K; int judge(LL x){
return W >= K*(get_all_S(x) - get_all_S(M-)) ;
} LL calc(){
LL L ,R ,mid , ans = -;
L = M ;
R = (LL) ;
while(L <= R){
mid = (L + R) >> ;
if(judge(mid)){
ans = mid ;
L = mid + ;
}
else
R = mid - ;
}
return ans==-?:ans - M + ;
} int main(){
init() ;
LL x ;
while(cin>>W>>M>>K){
cout<<calc()<<endl ;
}
return ;
}
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