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P3811 【模板】乘法逆元
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problem/P3811/P3811.cpp
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18
problem/P3811/P3811.cpp
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#include <bits/stdc++.h>
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using namespace std;
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long long inv[3000005];
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int main() {
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inv[0] = 0;
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inv[1] = 1;
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int n, p;
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scanf("%d%d", &n, &p);
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printf("1\n");
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for (int i = 2; i <= n; ++i) {
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inv[i] = (long long)p - (p / i) * inv[p % i] % p;
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printf("%d\n", inv[i]);
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}
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return 0;
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}
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