只能說自己以前做的那個矩陣快速冪太弱了,或者推的還不夠。
這道題把f(i),s(i),a(i),b(i)都合在一塊了,一個5*5的矩陣。
另外,反正在hdu就用%I64d,別管別的了。 keng...
我的paw,multiply模板肯定沒錯,連續的WA是坑在__int64上。
練習的題目太少,推敲涉及比較多的變量,贊一個
#include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __int64 LL; const LL maxn=1e18+7; const LL mod=1000000007; typedef struct M { LL x[6][6]; }mat; struct M multiply(struct M a,struct M b) { struct M ret; for(int i = 0;i<5;i++) { for(int j = 0;j <5;j++) { LL tem = 0; for(int k = 0;k < 5;k++) tem = (tem+a.x[i][k]*b.x[k][j])%mod; ret.x[i][j] = tem%mod;//這地方取模注意一下. } } return ret; } struct M paw(struct M a,long long t) { if(t==1) return a; else { struct M b=paw(a,t/2); if(t&1) { return multiply(multiply(b,b),a); } else return multiply(b,b); } } mat org,add; int main() { LL n,a0,ax,ay; LL b0,bx,by; while(scanf("%I64d",&n)!=EOF) { scanf("%I64d%I64d%I64d",&a0,&ax,&ay); scanf("%I64d%I64d%I64d",&b0,&bx,&by); if(n==0) {printf("0\n");continue;} memset(org.x,0,sizeof(org.x)); memset(add.x,0,sizeof(add.x)); add.x[0][0]=1; add.x[0][1]=0; add.x[0][2]=0; add.x[0][3]=0; add.x[0][4]=0; add.x[1][0]=1; add.x[1][1]=ax*bx%mod; add.x[1][2]=0; add.x[1][3]=0; add.x[1][4]=0; add.x[2][0]=0; add.x[2][1]=ax*by%mod; add.x[2][2]=ax%mod; add.x[2][3]=0; add.x[2][4]=0; add.x[3][0]=0; add.x[3][1]=ay*bx%mod; add.x[3][2]=0; add.x[3][3]=bx%mod; add.x[3][4]=0; add.x[4][0]=0; add.x[4][1]=ay*by%mod; add.x[4][2]=ay%mod; add.x[4][3]=by%mod; add.x[4][4]=1; org.x[0][0]=a0*b0%mod; org.x[0][1]=(((a0*ax+ay)%mod)*((b0*bx+by)%mod))%mod; org.x[0][2]=(a0*ax+ay)%mod; org.x[0][3]=(b0*bx+by)%mod; org.x[0][4]=1; mat ans; if(n==1) ans=org; else { ans=multiply(org,paw(add,n-1)); } printf("%I64d\n",ans.x[0][0]%mod); } return 0; } #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __int64 LL; const LL maxn=1e18+7; const LL mod=1000000007; typedef struct M { LL x[6][6]; }mat; struct M multiply(struct M a,struct M b) { struct M ret; for(int i = 0;i<5;i++) { for(int j = 0;j <5;j++) { LL tem = 0; for(int k = 0;k < 5;k++) tem = (tem+a.x[i][k]*b.x[k][j])%mod; ret.x[i][j] = tem%mod;//這地方取模注意一下. } } return ret; } struct M paw(struct M a,long long t) { if(t==1) return a; else { struct M b=paw(a,t/2); if(t&1) { return multiply(multiply(b,b),a); } else return multiply(b,b); } } mat org,add; int main() { LL n,a0,ax,ay; LL b0,bx,by; while(scanf("%I64d",&n)!=EOF) { scanf("%I64d%I64d%I64d",&a0,&ax,&ay); scanf("%I64d%I64d%I64d",&b0,&bx,&by); if(n==0) {printf("0\n");continue;} memset(org.x,0,sizeof(org.x)); memset(add.x,0,sizeof(add.x)); add.x[0][0]=1; add.x[0][1]=0; add.x[0][2]=0; add.x[0][3]=0; add.x[0][4]=0; add.x[1][0]=1; add.x[1][1]=ax*bx%mod; add.x[1][2]=0; add.x[1][3]=0; add.x[1][4]=0; add.x[2][0]=0; add.x[2][1]=ax*by%mod; add.x[2][2]=ax%mod; add.x[2][3]=0; add.x[2][4]=0; add.x[3][0]=0; add.x[3][1]=ay*bx%mod; add.x[3][2]=0; add.x[3][3]=bx%mod; add.x[3][4]=0; add.x[4][0]=0; add.x[4][1]=ay*by%mod; add.x[4][2]=ay%mod; add.x[4][3]=by%mod; add.x[4][4]=1; org.x[0][0]=a0*b0%mod; org.x[0][1]=(((a0*ax+ay)%mod)*((b0*bx+by)%mod))%mod; org.x[0][2]=(a0*ax+ay)%mod; org.x[0][3]=(b0*bx+by)%mod; org.x[0][4]=1; mat ans; if(n==1) ans=org; else { ans=multiply(org,paw(add,n-1)); } printf("%I64d\n",ans.x[0][0]%mod); } return 0; }