题解

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL P=200907;

LL qpow(LL a, LL b, LL c) {
    LL ret = 1 % c; a %= c;
    for (; b; b >>= 1, a = a * a % c) if (b & 1) ret = ret * a % c;
    return ret;
}

int main() {
    int T; scanf("%d", &T);
    while (T--) {
        LL a1, a2, a3, k; scanf("%lld%lld%lld%lld", &a1, &a2, &a3, &k);
        if (a2 - a1 == a3 - a2) { // 等差数列
            LL d = a2 - a1;
            printf("%lld\n", (a1 + (k - 1) * d % P) % P);
        } else { // 等比数列
            LL q = a2 / a1;
            printf("%lld\n", a1 * qpow(q, k - 1, P) % P);
        }
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL P=200907;
LL qpow(LL a,LL b,LL c)
{
    LL ret=1%c;a%=c;
    for(;b;b>>=1,a=a*a%c)if(b&1)ret=ret*a%c;
    return ret;
}
int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        LL a1,a2,a3,k;scanf("%lld%lld%lld%lld",&a1,&a2,&a3,&k);
        if(a2-a1==a3-a2)
        {
            LL d=a2-a1;
            printf("%lld\n",(a1+(k-1)*d%P)%P);
        }
        else
        {
            LL q=a2/a1;
            printf("%lld\n", a1*qpow(q,k-1,P)%P);
        }
    }
    return 0;
}