#include <stdio.h>
int n,m,d,dp[13][13],sum[13],dn[13],dm[13];
// DFS的时候这两个地方根据不同要求写。
int dfs(int da[], int dep, int all)
{
int i,j,ret=0;
if (dep == 0) return 1;
for (i=0;i<da[dep];i++)
{
if (all > 0 || i > 0) {
if (all == 0 || i-da[dep+1]>=d || i-da[dep+1]<=-d)
ret += dp[dep][i];
} else
ret += sum[dep-1];
}
if (all == 0 || da[dep]-da[dep+1]>=d || da[dep]-da[dep+1]<=-d)
ret += dfs(da, dep-1, all+da[dep]);
return ret;
}
int main()
{
int i,j,k,l,T;
scanf("%d", &T);
while (T--)
{
scanf("%d %d %d", &m, &n, &d);
for (i=0;i<13;i++)
for (j=0;j<13;j++) dp[i][j] = 0;
sum[0] = 0; sum[1] = 9;
for (i=0;i<10;i++) dp[1][i] = 1;
for (i=2;i<13;i++) {
sum[i] = sum[i-1];
for (j=0;j<10;j++) {
for (k=0;k<10;k++)
if (j-k>=d || j-k<=-d)
dp[i][j] += dp[i-1][k];
if (j > 0)
sum[i] += dp[i][j];
}
}
// for (i=0;i<=2;i++)
// for (j=0;j<10;j++) printf("%d %d %d\n", i, j, dp[i][j]);
i = 1; k = n;
while (i < 13) {
dn[i] = k % 10; k /= 10;
i++;
}
i = 1; k = m-1;
while (i < 13) {
dm[i] = k % 10; k /= 10;
i++;
}
n = dfs(dn, 11, 0);
if (m == 1)
m = 0;
else
m = dfs(dm, 11, 0);
printf("%d\n", n-m);
}
return 0;
}
Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 2997
Accepted: 1603
Description
A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?
Input
Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.
Output
For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.
Sample Input
10 11
100 200
0 500
1234567890 2345678901
0 4294967295
-1 -1
Sample Output
1
22
92
987654304
3825876150
Source
Waterloo Local Contest, 2006.5.27
import java.util.*;
import java.math.*;
import java.io.*;
public class Main {
static long val,n,m,dp[][]=new long[13][13],a[]=new long[13],dn[]=new long[13], dm[]=new long[13], sum[]=new long[13];
static long dfs(long dnm[], int dep, long all)
{
int i, j, k;
long ret=0;
if (dep == 0) return 0;
for (i=0;i<dnm[dep];i++) {
if (all > 0 || i > 0)
ret += dp[dep][i]; // 需要计算前导0
else
ret += sum[dep-1]; // 不需要计算前导0
}
if (all > 0 && dnm[dep] == 0)
ret += val % a[dep] + 1;
ret += dfs(dnm, dep-1, all+dnm[dep]);
return ret;
}
public static void main(String[] args) {
int i,j,k,l;
Scanner cin = new Scanner(System.in);
a[1] = 10;
for (i=2;i<13;i++) a[i] = a[i-1]*10;
for (i=0;i<13;i++)
for (j=0;j<13;j++) dp[i][j] = 0;
dp[1][0] = 1;
sum[0] = sum[1] = 0;
for (i=2;i<13;i++) {
sum[i] = sum[i-1];
for (j=0;j<10;j++) {
for (k=0;k<10;k++)
dp[i][j] += dp[i-1][k];
dp[i][j] += j==0 ? a[i-1] : 0;
if (j > 0)
sum[i] += dp[i][j];
}
}
while (true) {
m = cin.nextLong();
n = cin.nextLong();
if (m == -1 || n == -1) break;
for (i=0;i<13;i++) dn[i] = dm[i] = 0;
i = 1;
val = n;
while (val > 0) {
dn[i] = val % 10;
val /= 10;
i++;
}
i = 1;
val = m-1;
while (val > 0) {
dm[i] = val % 10;
val /= 10;
i++;
}
val = n;
n = dfs(dn, 12, 0) + 1; // 0 还有一个0
val = m-1;
m = dfs(dm, 12, 0) + 1;
if (val < 0) m = 0;
System.out.println(n-m);
}
}
}
$ /usr/lib/rpm/debugedit
Usage: debugedit [OPTION...]
-b, --base-dir=STRING base build directory of objects
-d, --dest-dir=STRING directory to rewrite base-dir into
-l, --list-file=STRING file where to put list of source and header file
names
-i, --build-id recompute build ID note and print ID on stdout
Help options:
-?, --help Show this help message
--usage Display brief usage message
