HDU 2254 奥运（矩阵快速幂+二分等比序列求和）

ACM

题目地址：HDU 2254 奥运

题意： 
中问题不解释。

分析： 
根据floyd的算法，矩阵的k次方表示这个矩阵走了k步。 
所以k天后就算矩阵的k次方。 
这样就变成：初始矩阵的^[t1,t2]这个区间内的v[v1][v2]的和。 
所以就是二分等比序列求和上场的时候了。 
跟HDU 1588 Gauss Fibonacci的算法一样。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        2254.cpp*  Create Date: 2014-08-04 10:52:29*  Descripton:  matrix, floyd*/#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 20;const int SIZE = 32;        // max size of the matrixconst int MOD = 2008;int n, k, p1, p2;ll v1, v2, t1, t2;map<int, int> mp;struct Mat{    int n;    ll v[SIZE][SIZE];    // value of matrix    Mat(int _n = SIZE) {        n = _n;    }    void init(ll _v = 0) {        memset(v, 0, sizeof(v));        if (_v)            repf (i, 0, n - 1)                v[i][i] = _v;    }    void output() {        repf (i, 0, n - 1) {            repf (j, 0, n - 1)                printf("%lld ", v[i][j]);            puts("");        }        puts("");    }} a, b;Mat operator * (Mat a, Mat b) {    Mat c(a.n);    repf (i, 0, a.n - 1) {        repf (j, 0, a.n - 1) {            c.v[i][j] = 0;            repf (k, 0, a.n - 1) {                c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;                c.v[i][j] %= MOD;            }        }    }    return c;}Mat operator ^ (Mat a, ll k) {    Mat c(a.n);    c.init(1);    while (k) {        if (k&1) c = a * c;        a = a * a;        k >>= 1;    }    return c;}Mat operator + (Mat a, Mat b) {    Mat c(a.n);    repf (i, 0, a.n - 1)        repf (j, 0, a.n - 1)            c.v[i][j] = (b.v[i][j] + a.v[i][j]) % MOD;    return c;}Mat operator + (Mat a, ll b) {    Mat c = a;    repf (i, 0, a.n - 1)        c.v[i][i] = (a.v[i][i] + b) % MOD;    return c;}Mat calc(Mat a, int n) {    if (n == 1)        return a;    if (n&1)        return (a^n) + calc(a, n - 1);    else        return calc(a, n/2) * ((a^(n/2)) + 1);}int main() {    while (~scanf("%d", &n)) {        a.init();        mp.clear();        int cnt = 0;        while (n--) {            scanf("%d%d", &p1, &p2);            if (mp.find(p1) == mp.end())                p1 = mp[p1] = cnt++;            else                p1 = mp[p1];            if (mp.find(p2) == mp.end())                p2 = mp[p2] = cnt++;            else                p2 = mp[p2];            a.v[p1][p2]++;        }        a.n = cnt;        scanf("%d", &k);        while (k--) {            scanf("%lld%lld%lld%lld", &v1, &v2, &t1, &t2);            if (mp.find(v1) == mp.end() || mp.find(v2) == mp.end()) {                puts("0");                continue;            }            v1 = mp[v1];            v2 = mp[v2];            if (t1 > t2)                swap(t1, t2);            if (t1 == 0) {                if (t2 == 0)                    puts("0");                else                    printf("%lld\n", calc(a, t2).v[v1][v2]);            }            else if (t1 == 1)                printf("%lld\n", calc(a, t2).v[v1][v2]);            else {                printf("%lld\n", ((calc(a, t2).v[v1][v2] - calc(a, t1 - 1).v[v1][v2]) + MOD) % MOD);            }        }    }    return 0;}