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TangDou/LanQiaoBei/LanQiao14STEMA202211/4_1.cpp
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30
TangDou/LanQiaoBei/LanQiao14STEMA202211/4_1.cpp
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#include <bits/stdc++.h>
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/*
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测试用例I:
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3 7
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答案:5
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测试用例II:
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4 15
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答案:144
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测试用例III:
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1 22
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答案:17711
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*/
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using namespace std;
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int n, m;
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int f[110];
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int main() {
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cin >> n >> m;
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f[2] = 1, f[3] = 2;
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for (int i = 4; i <= m - n + 1; i++) f[i] = f[i - 1] + f[i - 2];
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printf("%d\n", f[m - n + 1]);
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return 0;
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}
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