點擊此處即可傳送HIT 2060

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 mod 1,000,000,000, since S may be quite large. Sample Input 1 1 3 5 10 1000 0 0Sample Output 1 16 496035733

題目大意：就是給你一個a和b， 求從a到b之間的斐波那契數的和

解題思路：矩陣乘法，注意考慮邊界條件：
具體詳見代碼：
上代碼：

/* 2015 - 8 - 14 上午 Author: ITAK 今日的我要超越昨日的我，明日的我要勝過今日的我， 以創作出更好的代碼為目標，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);}}else if(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);}}return 0; }

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