POJ_2282 The Counting Problem
题目链接：http://poj.org/problem?id=2282
暴力就不说了，对于求0~n所有数字中0~9出现过的次数来说，可以直接对n的每一位计算出0~9中每个数字出现的次数（数位dp还不会。。。。）。
仔细观察能够发现，例如n=2406， 那么对于百位来说：
0 2*100次
1 2*100+100
2 2*100+100
3 2*100+100
4 2*100+7
5 2*100
6 2*100
7 2*100
8 2*100
9 2*100
也就是说数n每一位上0~9出现的次数与该位之前的数以及该位后面的数有关。
细节较多，需要仔细考虑好。
代码：
1 long long cnt1[11], cnt2[11]; 2 3 int dowork(long long n, long long cnt[]){ 4 cnt[0]++; 5 long long tmx = n; 6 for(int i = 0; tmx; i++){ 7 long long ladigit = tmx % 10; 8 tmx /= 10; 9 for(int j = 0; j < 10; j++){ 10 if(!tmx && !j) 11 continue; 12 if(j < ladigit){ 13 if(!tmx) cnt[j] += 1 * pow(10, i); 14 else cnt[j] += tmx * pow(10, i); 15 if(j && tmx) cnt[j] += pow(10, i); 16 } 17 else if(j == ladigit){ 18 if(!tmx) cnt[j] += n - j * pow(10, i) + 1; 19 else if(j) cnt[j] += tmx * pow(10, i) + n - tmx * pow(10, i + 1) - j * pow(10, i) + 1; 20 else cnt[j] += (tmx - 1) * pow(10, i) + n - tmx * pow(10, i + 1) - j * pow(10, i) + 1; 21 22 } 23 else{ 24 cnt[j] += tmx * pow(10, i); 25 } 26 } 27 /* 28 printf("tmp-->"); 29 for(int i = 0; i < 10; i++) 30 printf("%d ", cnt[i]); 31 puts(""); 32 */ 33 } 34 return 0; 35 } 36 37 int main(){ 38 long long n1, n2; 39 while(scanf("%lld %lld", &n1, &n2) && n1){ 40 memset(cnt1, 0, sizeof(cnt1)); 41 memset(cnt2, 0, sizeof(cnt2)); 42 dowork(min(n1, n2) - 1, cnt1); 43 dowork(max(n1, n2), cnt2); 44 for(int i = 0; i < 10; i++) 45 printf("%lld ", cnt2[i] - cnt1[i]); 46 puts(""); 47 } 48 }
题目：
The Counting Problem
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 4561 | Accepted: 2342 |
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.
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
