題目鏈接

Problem Description

We all know the definition of Fibonacci series: fib[i]=fib[i-1]+fib[i-2],fib[1]=1,fib[2]=1.And we define another series P associated with the Fibonacci series: P[i]=fib[4*i-1].Now we will give several queries about P:give two integers L,R, and calculate ∑P[i](L <= i <= R).

Input

There is only one test case.The first line contains single integer Q – the number of queries. (Q<=10^4)Each line next will contain two integer L, R. (1<=L<=R<=10^12)

Output

For each query output one line.Due to the final answer would be so large, please output the answer mod 1000000007.

Sample Input

2
1 300
2 400

Sample Output

838985007
352105429

AC

• 經過以下推導，只用求單個斐波那契值即可，因為n比較大，用矩陣快速冪

#include <iostream> #include <algorithm> #include <stdio.h> #include <vector> #include <map> #include <bitset> #include <set> #include <string.h> #include <cmath> #include <queue> #include <algorithm> #define N 100005 #define P pair<int,int> #define ll long long #define lowbit(a) a&(-a) #define mk(a, b) make_pair(a, b) #define mem(a, b) memset(a, b, sizeof(a)) ll mod = 1e9 +7; using namespace std; // 矩陣相乘 void mul(ll a[][2], ll b[][2]) {ll ans[2][2];for (int i = 0; i < 2; ++i) {for (int j = 0; j < 2; ++j) {ans[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];}}a[0][0] = ans[0][0] % mod;a[0][1] = ans[0][1] % mod;a[1][0] = ans[1][0] % mod;a[1][1] = ans[1][1] % mod; } ll x[2][2], t[2][2]; // 矩陣快速冪 void quick(ll num) {x[0][0] = 0;x[0][1] = 1;t[0][0] = 0;t[0][1] = 1;t[1][0] = 1;t[1][1] = 1;while (num) {if (num & 1) {mul(x, t);}num >>= 1;mul(t, t);} } int main(){ #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout); #endifint t;scanf("%d", &t);while (t--) {ll l, r;scanf("%lld%lld", &l, &r);l--;ll sumr, suml;quick(2 * r);sumr = x[0][0] * x[0][1] % mod;quick(2 * l);suml = x[0][0] * x[0][1] % mod;ll ans = sumr - suml;printf("%lld\n", (ans + mod) % mod);}#ifndef ONLINE_JUDGEfclose(stdin);fclose(stdout);system("out.txt"); #endifreturn 0; }

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