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#include <bits/stdc++.h>
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using namespace std;
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const int N = 10;
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// 暴力法获取从1开始到n,有多少个指定的x,[类似于前缀和的思路]
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// 从0到10的8次方,就是枚举每一位,一个测试点是 8*10^8,会超时,不可取
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int force_count(int n, int x) {
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int res = 0;
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for (int i = 1; i <= n; i++) {
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int t = i;
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while (t) {
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if (t % 10 == x) res++;
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t /= 10;
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}
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}
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return res;
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}
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int main() {
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int a, b;
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// 当读入一行为0 0时,表示输入终止,且该行不作处理,注意这里 a||b的使用方法
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while (cin >> a >> b, a || b) {
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if (a > b) swap(a, b); // 这题还玩小的在后,大的在前,需要我们用代码判断,shit!
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// 计算0--9的每一个数出现的次数
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for (int i = 0; i <= 9; i++)
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cout << force_count(b, i) - force_count(a - 1, i) << ' ';
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cout << endl;
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}
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return 0;
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}
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