标签:style blog http color strong os
原题:ZOJ 3772 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3772
这题算是长见识了,还从没坐过矩阵+线段树的题目呢,不要以为矩阵就一定配合快速幂来解递推式的哦。
由F(x)=F(x-1)+F(x-2)*A[x],转化为矩阵乘法:
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所以维护一颗线段树,线段树的每个结点保存一个矩阵,叶子节点为: a[0][0] = a[1][0] = 1, a[0][1] = Ax, a[1][1] = 0的形式
否则保存 a[r] * a[r-1] * ... * a[l] 的结果矩阵,且要注意不要乘反了。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #define Mod 1000000007 #define ll long long using namespace std; #define N 100007 struct Matrix { ll m[2][2]; Matrix(int _x) { m[0][0] = 1; m[0][1] = _x; m[1][0] = 1; m[1][1] = 0; } Matrix(){} }tree[4*N]; ll A[N]; Matrix Mul(Matrix a,Matrix b) { Matrix c; memset(c.m,0,sizeof(c.m)); for(int i=0;i<2;i++) for(int j=0;j<2;j++) for(int k=0;k<2;k++) c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j])%Mod; return c; } void build(int l,int r,int rt) { if(l == r) { tree[rt] = Matrix(A[l]); return; } int mid = (l+r)/2; build(l,mid,2*rt); build(mid+1,r,2*rt+1); tree[rt] = Mul(tree[2*rt+1],tree[2*rt]); //注意不要乘反了 } Matrix query(int l,int r,int aa,int bb,int rt) { if(aa<=l && bb>=r) return tree[rt]; int mid = (l+r)/2; if(aa>mid) return query(mid+1,r,aa,bb,2*rt+1); else if(bb<=mid) return query(l,mid,aa,bb,2*rt); else return Mul(query(mid+1,r,aa,bb,2*rt+1),query(l,mid,aa,bb,2*rt)); //不要乘反 } int main() { int n,m; int i,t; int aa,bb; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=1;i<=n;i++) scanf("%lld",&A[i]); build(1,n,1); while(m--) { scanf("%d%d",&aa,&bb); if(bb-aa<=1) printf("%lld\n",A[bb]); else { Matrix ans = query(1,n,aa+2,bb,1); ll res = ans.m[0][0]*A[aa+1]%Mod + ans.m[0][1]*A[aa]%Mod; printf("%lld\n",res%Mod); } } } return 0; }
2014 Super Training #7 E Calculate the Function --矩阵+线段树,布布扣,bubuko.com
2014 Super Training #7 E Calculate the Function --矩阵+线段树
标签:style blog http color strong os
原文地址:http://www.cnblogs.com/whatbeg/p/3834569.html