Ultraviolet A 10692-Huge Mod (number theory)

Link to the question: Ultraviolet A 10692-Huge mod

Give the power of a number, and then it modulo the M value.

Solution: based on the nature of the residual series, the last line must be cyclical, so there will be AB = abmod (phi [M]) + phi [M] (phi [M] is an Euler's function of M), which can be solved based on recursion.

#include 
  
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    #include 
    
     const int maxn = 15;int A[maxn], k;int pow_mod (int a, int n, int M) {    int ans = 1;    while (n) {        if (n&1)            ans = ans * a % M;        a = a * a % M;        n /= 2;    }    return ans;}int euler_phi(int n) {    int m = (int)sqrt(n+0.5);    int ans = n;    for (int i = 2; i <= m; i++) {        if (n % i == 0) {            ans = ans / i * (i-1);            while (n%i==0)                n /= i;        }    }    if (n > 1)        ans = ans / n * (n - 1);    return ans;}int solve (int d, int M) {    if (d == k - 1)        return A[d]%M;    int phi = euler_phi(M);        int c = solve (d+1, phi) + phi;    return pow_mod(A[d], c, M);}int main () {    int cas = 1;    char str[maxn];    while (scanf("%s", str) == 1 && strcmp(str, "#")) {        int M;        sscanf(str, "%d", &M);        scanf("%d", &k);        for (int i = 0; i < k; i++)            scanf("%d", &A[i]);        printf("Case #%d: %d\n", cas++, solve(0, M));    }    return 0;}