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| Time Limit: 1000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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
分析：题目要求当我们写下m, m+1, ..., n时，总共写下了多少个0
所以，枚举0出现在每一位上时的情况数，将全部情况数汇总求和，即可不多不少地计算总数。
这里要注意，我们写数字时不可能写出一个数字是0开头的（不允许前导0），这一点留心。
#include <cstdio> #include <iostream> using namespace std; typedef long long LL; int base[11]; void Init() { base[1] = 1; for(int i=2; i<=10; i++) { base[i] = base[i-1] * 10; } } LL f(LL x) { if(x == -1) return 0; LL ret = 1; // 0的情况，下面没有讨论，在此补上 LL temp = x; for(int i=1; temp/10; i++, temp/=10) { // 因为最高位不用(不允许前导0)，所以判断条件写为 temp/10 int now = temp % 10; LL l = x / base[i+1]; if(now) { ret += l * base[i]; } else { LL r = x % base[i]; ret += (l-1) * base[i] + r + 1LL; } } return ret; } int main () { Init(); LL m, n; while(scanf("%lld%lld", &m, &n) != EOF && m!=-1) { printf("%lld\n", f(n) - f(m-1)); } return 0; }
POJ - 3286 - How many 0's? 【数位DP】
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原文地址：http://www.cnblogs.com/AcIsFun/p/5428661.html
