zoj 3772 Calculate the Function 线段树+矩阵

Time Limit: 2 Seconds Memory Limit: 65536 KB

You are given a list of numbers A1 A2 .. AN and M queries. For the i-th query:

The query has two parameters Li and Ri.
The query will define a function Fi(x) on the domain [Li, Ri] ∈ Z.
Fi(Li) = ALi
Fi(Li + 1) = A(Li + 1)
for all x >= Li + 2, Fi(x) = Fi(x - 1) + Fi(x - 2) × Ax

You task is to calculate Fi(Ri) for each query. Because the answer can be very large, you should output the remainder of the answer divided by 1000000007.
Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

The first line contains two integers N, M (1 <= N, M <= 100000). The second line contains N integers A1 A2 .. AN (1 <= Ai <= 1000000000).

The next M lines, each line is a query with two integer parameters Li, Ri (1 <= Li <= Ri <= N).
Output

For each test case, output the remainder of the answer divided by 1000000007.
Sample Input

1
4 7
1 2 3 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4

Sample Output

1
2
5
13
11
4

4

#include <stdio.h>#include <stdlib.h>#include <iostream>#define ls(x) ((x) << 1)#define rs(x) ((x) << 1 | 1)using namespace std;typedef long long ll;struct Mat {    int m[2][2];    void setE() {        m[0][0] = m[1][1] = 1;        m[0][1] = m[1][0] = 0;    }    void setZ() {        m[0][0] = m[0][1] = m[1][0] = m[1][1] = 0;    }};const int maxn = 100000 + 10;const int mod = 1000000007;int a[maxn];Mat mat[maxn * 8];Mat mul(Mat a, Mat b) {    Mat ret;    ret.setZ();    for (int i = 0; i < 2; i++)        for (int j = 0; j < 2; j++)            for (int k = 0; k < 2; k++)                ret.m[i][j] = ((ll)a.m[i][k] * b.m[k][j] + ret.m[i][j]) % mod;    return ret;}void Build(int cur, int  l, int r) {    int m;    if(l == r) {        mat[cur].m[0][0] = 0;        mat[cur].m[0][1] = 1;        mat[cur].m[1][0] = a[l];        mat[cur].m[1][1] = 1;    } else {        m = l + ((r - l) >> 1);        Build(ls(cur), l, m);        Build(rs(cur), m + 1, r);        mat[cur] = mul(mat[rs(cur)], mat[ls(cur)]);    }}Mat Query(int cur, int l, int r, int ql, int qr) {    int m;    Mat ret;    if (ql <= l && r <= qr)        return mat[cur];    else {        m = l + ((r - l) >> 1);        ret.setE();        if (ql <= m)            ret = mul(ret, Query(ls(cur), l, m, ql, qr));        if (m < qr)            ret = mul(Query(rs(cur), m + 1, r, ql, qr), ret);    }    return ret;}int main() {//  freopen("data.in", "r", stdin);    int tcas, n, m, l, r;    Mat L, R, ret;    scanf("%d", &tcas);    while(tcas--) {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++)            scanf("%d", &a[i]);        Build(1, 1, n);        while (m--) {            scanf("%d%d", &l, &r);            R.m[0][0] = a[l];            R.m[0][1] = 0;            R.m[1][0] = a[l+1];            R.m[1][1] = 0;            if (r - l <= 1)                printf("%d\n", a[r]);            else {                L = Query(1, 1, n, l + 2, r);                ret = mul(L, R);                printf("%d\n", ret.m[1][0]);            }        }    }    return 0;}