UVA 1640(数位统计)
来源：程序员人生 发布时间：2015-05-28 09:21:47 阅读次数：2729次
The Counting Problem
| Time Limit:3000MS |
|
Memory Limit:Unknown |
|
64bit IO Format:%lld & %llu |
SubmitStatus
Description
Given two integers a and
b, we write the numbers between a and
b, inclusive, in a list. Your task is to calculate the number of occurrences of each digit. For example, if
a = 1024 and b = 1032, the list will be
1024 1025 1026 1027 1028 1029 1030 1031 1032
there are ten 0's in the list, ten 1's, seven 2's, three 3's, and etc.
Input
The input consists of up to 500 lines. Each line contains two numbers
a and b where
0 < a, b < 100000000. The input is terminated by a line `0 0', which is not considered as part of the input.
Output
For each pair of input, output a line containing ten numbers separated by single spaces. The first number is the number of occurrences of the digit 0, the second is the number of occurrences of the digit 1, etc.
Sample Input
1 10
44 497
346 542
1199 1748
1496 1403
1004 503
1714 190
1317 854
1976 494
1001 1960
0 0
Sample Output
1 2 1 1 1 1 1 1 1 1
85 185 185 185 190 96 96 96 95 93
40 40 40 93 136 82 40 40 40 40
115 666 215 215 214 205 205 154 105 106
16 113 19 20 114 20 20 19 19 16
107 105 100 101 101 197 200 200 200 200
413 1133 503 503 503 502 502 417 402 412
196 512 186 104 87 93 97 97 142 196
398 1375 398 398 405 499 499 495 488 471
294 1256 296 296 296 296 287 286 286 247
Source
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths ::
Examples
Submit
Status
统计区间里0到9每一个数字出现了多少次，直接数学瞎弄
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10 + 10;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
typedef long long ll;
int c[maxn][maxn];
void init(int n)
{
for(int i = 0; i <= n; i++) {
c[i][0] = 1;
for(int j = 1; j <= i; j++) c[i][j] = c[i⑴][j⑴] + c[i⑴][j];
c[i][i+1] = 0;
}
}
ll quick(int a,int n)
{
ll res = 1,b = a;
while(n>0) {
if(n&1)res *= b;
b = b*b;
n>>=1;
}
return res;
}
ll work(int f,int k,int L,int tot)/*计算以f开头长为L且之前已出现tot个k的数字里有多少个k*/
{
ll res = 0;
if(f==k)tot++;
for(int i = 0; i < L; i++) {
res += (i+tot)*c[L⑴][i]*quick(9,L⑴-i);
}
return res;
}
ll solve(const char *s,int k)/*计算[1,s)里的数字1共有多少个k*/
{
int sz = strlen(s);
ll res = 0,tot = 0;
for(int i = 1; i < s[0]-'0'; i++)res += work(i,k,sz,0);
for(int L = 1; L < sz; L++) {
for(int i = 1; i < 10; i++)
res += work(i,k,L,0);
}
if(s[0]-'0'==k)tot++;
for(int i = 1; i < sz; i++) {
int limt = s[i] - '0';
for(int j = 0; j < limt; j++) {
res += work(j,k,sz-i,tot);
}
if(limt==k)tot++;
}
return res;
}
int main()
{
init(12);
int L,R;
while(~scanf("%d%d",&L,&R)) {
if(L+R==0)return 0;
if(L>R)swap(L,R);
char sl[20],sr[20];
sprintf(sl,"%d",L);
sprintf(sr,"%d",R+1);
for(int i = 0; i < 10; i++)
printf("%lld%c",solve(sr,i)-solve(sl,i),"
"[i==9]);
}
return 0;
}
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