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Description
Problem E: How many 0's?
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 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.
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
Output for sample input
1 22 92 987654304 3825876150 题意： 求[n, m]之间包含0的数字的个数 思路：先转化为前缀问题，就是cal(m)-cal(n-1)，枚举每一位，如果第i位是0的话，那么为了不让数大于原本的数的话，那么它的左边取[1, a-1]的时候,右边可以取任意值 如果取a的话，那么只能取0到右边的最大；如果是非0的话，那么让这位为0的话，就可以左边随便取，右边随便取#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> typedef long long ll; using namespace std;ll n, m;ll cal(ll x) {if (x < 0)return 0;ll ans = 1, cnt = 1, tmp = 0;while (x >= 10) {ll cur = x % 10;x /= 10;if (cur) ans += x * cnt;else ans += (x-1) * cnt + tmp + 1;tmp += cur * cnt;cnt *= 10;}return ans; }int main() {while (scanf("%lld%lld", &n, &m) != EOF && n != -1 && m != -1) {printf("%lld\n", cal(m) - cal(n-1)); } return 0; }
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