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40 lines
966 B
40 lines
966 B
#include <bits/stdc++.h>
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using namespace std;
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const int N = 32;
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int l, r;
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int a[N], al;
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int f[N][N];
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/**
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*
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* @param u 当前的数位
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* @param st 前一位填写的是什么
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* @param op 是不是贴上界
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* @return 在当前情况下，后续符合条件的有多少个
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*/
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int dfs(int u, int st, bool op) {
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if (u == 0) return 1; // 如果走完所有情况，还活着，说明前面的都符合条件，就是一个合理解++
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if (!op && ~f[u][st]) return f[u][st];
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int ans = 0, up = op ? a[u] : 9;
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for (int i = st; i <= up; i++) // 注意起始点,i >= st
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ans += dfs(u - 1, i, op && i == a[u]);
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if (!op) f[u][st] = ans;
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return ans;
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}
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int calc(int x) {
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al = 0;
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while (x) a[++al] = x % 10, x /= 10;
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return dfs(al, 0, true);
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}
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int main() {
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memset(f, -1, sizeof f);
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while (~scanf("%d%d", &l, &r))
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printf("%d\n", calc(r) - calc(l - 1));
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return 0;
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}
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