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#include <bits/stdc++.h>
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using namespace std;
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// 如果 l,r之间的跨度大于等于一个n, 那么我们肯定可以想办法拿到n-1
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// 如果l,r之间的跨度小于一个n,那么r%n就是答案
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int main() {
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int n, l, r;
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cin >> n >> l >> r;
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if (l / n == r / n)
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cout << r % n << endl;
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else
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cout << n - 1 << endl;
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return 0;
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} |
