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https://github.com/Kiritow/OJ-Problems-Source.git
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83 lines
2.5 KiB
C++
83 lines
2.5 KiB
C++
#include<cstdio>
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using namespace std;
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typedef long long ll;
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const ll mod = 2147493647;
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ll n,a,b;
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struct Matrix
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{
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ll m[7][7];
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void init1()
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{
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m[0][0] = b,m[0][1] = 0,m[0][2] = 0,m[0][3] = 0,m[0][4] = 0,m[0][5] = 0,m[0][6] = 0;
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m[1][0] = a,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0;
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m[2][0] = 16,m[2][1] = 0,m[2][2] = 0,m[2][3] = 0,m[2][4] = 0,m[2][5] = 0,m[2][6] = 0;
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m[3][0] = 8,m[3][1] = 0,m[3][2] = 0,m[3][3] = 0,m[3][4] = 0,m[3][5] = 0,m[3][6] = 0;
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m[4][0] = 4,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 0,m[4][5] = 0,m[4][6] = 0;
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m[5][0] = 2,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 0,m[5][6] = 0;
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m[6][0] = 1,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 0;
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}
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void init2()
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{
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m[0][0] = 1,m[0][1] = 2,m[0][2] = 1,m[0][3] = 4,m[0][4] = 6,m[0][5] = 4,m[0][6] = 1;
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m[1][0] = 1,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0;
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m[2][0] = 0,m[2][1] = 0,m[2][2] = 1,m[2][3] = 4,m[2][4] = 6,m[2][5] = 4,m[2][6] = 1;
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m[3][0] = 0,m[3][1] = 0,m[3][2] = 0,m[3][3] = 1,m[3][4] = 3,m[3][5] = 3,m[3][6] = 1;
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m[4][0] = 0,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 1,m[4][5] = 2,m[4][6] = 1;
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m[5][0] = 0,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 1,m[5][6] = 1;
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m[6][0] = 0,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 1;
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}
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Matrix operator * (Matrix t)
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{
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Matrix res;
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for (int i = 0; i < 7; i++)
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{
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for (int j = 0; j < 7; j++)
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{
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res.m[i][j] = 0;
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for (int k = 0;k < 7; k++)
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res.m[i][j] = (res.m[i][j] + (m[i][k] % mod) * (t.m[k][j] % mod) % mod) % mod;
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}
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}
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return res;
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}
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Matrix operator ^ (int k)
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{
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Matrix res,s;
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res.init2();
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s.init2();
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while(k)
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{
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if(k & 1)
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res = res * s;
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k >>= 1;
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s = s * s;
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}
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return res;
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}
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};
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int main()
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{
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int T;
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scanf("%d",&T);
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while(T--)
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{
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scanf("%lld %lld %lld",&n,&a,&b);
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if(n == 1)
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{
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printf("%lld\n",a % mod);
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continue;
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}
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if(n == 2)
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{
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printf("%lld\n",b % mod);
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continue;
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}
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Matrix ans,t;
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ans.init1();
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t.init2();
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ans = (t^(n-3)) * ans;
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printf("%lld\n",ans.m[0][0]);
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}
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return 0;
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}
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