【poj3070】Fibonacci (矩阵乘法+快速幂)

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3.16

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[2][2],b[2][2],n;void mul(int a[2][2],int b[2][2],int ans[2][2]){int t[2][2];memset(t,0,sizeof(t));for(int i=0;i<=1;i++)for(int j=0;j<=1;j++)for(int k=0;k<=1;k++)t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000;for(int i=0;i<=1;i++)for(int j=0;j<=1;j++)ans[i][j]=t[i][j];}int main(){while(scanf("%d",&n)!=EOF){a[0][0]=a[0][1]=a[1][0]=b[0][0]=b[1][1]=1;        b[1][0]=b[0][1]=a[1][1]=0;if(n==-1)return 0;while(n){if(n&1)mul(a,b,b);//b=a*b;n>>=1;//n/=2;mul(a,a,a);//a*=a;}printf("%d\n",b[1][0]);}}

8.12

#include<iostream>#include<cstdio>using namespace std;inline int read() {    int x = 0, f = 1;    char ch = getchar();    while (ch < '0' || ch > '9') {        if (ch == '-')f = -1;        ch = getchar();    }    while (ch >= '0' && ch <= '9') {        x = x * 10 + ch - '0';        ch = getchar();    }    return x*f;}struct M {    int a[2][2];} a;inline M operator *(const M &a, const M &b) {    M res;    for (int i = 0; i < 2; ++i)        for (int j = 0; j < 2; ++j) {            res.a[i][j] = 0;            for (int k = 0; k < 2; ++k)                res.a[i][j] = (res.a[i][j] + a.a[i][k] * b.a[k][j]) % 10000;        }    return res;}inline M operator ^ (M a, int n) {    M res = a;    for (--n; n; n >>= 1, a = a * a)        if (n & 1)res = res * a;    return res;}int n;int main() {    while (~scanf("%d", &n)) {        if (n == -1)return 0;        if (n == 0) {            printf("0\n");            continue;        }        a.a[0][0] = a.a[0][1] = a.a[1][0] = 1;        a.a[1][1] = 0;        a = a^n;        printf("%d\n", a.a[1][0]);    }}

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