As we know , the Fibonacci numbers are defined as follows: F(n) == {1 n==0||n==1{F(n-1)+F(n-2) n>1;
Given two numbers a and b , calculate . 從a到b之間的斐波那契數(shù)的和Input The input contains several test cases. Each test case consists of two non-negative integer numbers a and b (0 ≤ a ≤ b ≤1,000,000,000). Input is terminated by a = b = 0. Output For each test case, output S mod1,000,000,000, since S may be quite large. Sample Input
113510100000Sample Output 116496035733
題目大意：就是給你一個(gè)a和b， 求從a到b之間的斐波那契數(shù)的和
解題思路：矩陣乘法，注意考慮邊界條件： 具體詳見(jiàn)代碼： 上代碼：
/*
2015 - 8 - 14 上午
Author: ITAK
今日的我要超越昨日的我，明日的我要?jiǎng)龠^(guò)今日的我，
以創(chuàng)作出更好的代碼為目標(biāo)，不斷地超越自己。
*/#include <iostream>#include <cstdio>using namespace std;
const int maxn = 3;
typedef long long LL;
const int mod = 1e9;
typedef struct
{LL m[maxn][maxn];
} Matrix;Matrix P = {1,1,0,1,0,0,1,1,1,};
Matrix I = {1,0,0,0,1,0,0,0,1,};
Matrix matrix_mul(Matrix a, Matrix b)
{int i, j, k;Matrix c;for(i=0; i<maxn; i++){for(j=0; j<maxn; j++){c.m[i][j] = 0;for(k=0; k<maxn; k++)c.m[i][j] += (a.m[i][k] * b.m[k][j]) % mod;c.m[i][j] %= mod;}}return c;
}Matrix quick_mod(LL m)
{Matrix ans = I, b = P;while(m){if(m & 1)ans = matrix_mul(ans, b);m >>= 1;b = matrix_mul(b, b);}return ans;
}
int main()
{LL a, b;while(~scanf("%lld%lld",&a,&b)){Matrix tmp1, tmp2;if(!a && !b)break;if(a == 0){if(b == 1)puts("2");else{tmp2 = quick_mod(b-1);LL ans2 = tmp2.m[2][0]+tmp2.m[2][1]+2*tmp2.m[2][2];LL ans = (ans2%mod-1)%mod;if(ans < 0)ans += mod;printf("%lld\n",ans+1);}}elseif(a == 1){if(b == 1)puts("1");else{tmp2 = quick_mod(b-1);LL ans2 = tmp2.m[2][0]+tmp2.m[2][1]+2*tmp2.m[2][2];LL ans = (ans2%mod-1)%mod;if(ans < 0)ans += mod;printf("%lld\n",ans);}}else{tmp1 = quick_mod(a-2);tmp2 = quick_mod(b-1);LL ans1 = tmp1.m[2][0]+tmp1.m[2][1]+2*tmp1.m[2][2];LL ans2 = tmp2.m[2][0]+tmp2.m[2][1]+2*tmp2.m[2][2];//cout<<ans1 << " "<<ans2<<" ";LL ans = (ans2%mod - ans1%mod);if(ans < 0)ans += mod;printf("%lld\n",ans);}}return0;
}
