Day7 - J - Raising Modulo Numbers POJ - 1995
Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions Ai Bi from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers.
You should write a program that calculates the result and is able to find out who won the game.
Input
Output
(A1B1+A2B2+ ... +AHBH)mod M.
Sample Input
3
16
4
2 3
3 4
4 5
5 6
36123
1
2374859 3029382
17
1
3 18132
Sample Output
2
13195
13
思路:题目一大串,有用的就是output(, 裸的欧拉降幂,直接打表求欧拉函数直接算即可(扩展欧拉)
typedef long long LL;
typedef pair<LL, LL> PLL; const int maxm = ; int phi[maxm]; void getEuler() {
phi[] = ;
for(int i = ; i < maxm; ++i) {
if(!phi[i]) {
for(int j = i; j < maxm; j += i) {
if(!phi[j]) phi[j] = j;
phi[j] = phi[j] / i * (i - );
}
}
}
} LL quick_pow(LL a, LL b, LL p) { //a^b(modp)
LL ret = ;
while(b) {
if(b&) ret = (ret * a) % p;
a = (a * a) % p;
b >>= ;
}
return ret;
} int main() {
int T, H;
getEuler();
LL MOD, A, B;
scanf("%d", &T);
while(T--) {
scanf("%lld", &MOD);
LL ans = ;
scanf("%d", &H);
for(int i = ; i < H; ++i) {
scanf("%lld%lld", &A, &B);
if(B < phi[MOD])
ans = (ans + quick_pow(A, B, MOD)) % MOD;
else
ans = (ans + quick_pow(A, B%phi[MOD]+phi[MOD], MOD)) % MOD;
}
printf("%lld\n", ans);
}
return ;
}
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