HDU 4920 Matrix multiplication

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2014 Multi-University Training Contest 5

其实根本不是一题水题，具体题解见HDU微博

但是因为叉姐没有卡读入挂，你加个读入挂就可以过了

或者，你用C++的编译语言交也可以过

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>using namespace std;#define L 805template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}int n;int ta[L];int tb[L];int a[L][L];int b[L][L];int c[L][L];void matrixMul(){    memset(c,0,sizeof(c));    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            if(a[i][j]==0) continue;            for(int k=1;k<=n;k++) c[i][k] = (c[i][k] + a[i][j]*b[j][k])%3;        }    }}int main(){    while(read(n)){        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                read(a[i][j]),a[i][j]%=3;        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                read(b[i][j]),b[i][j]%=3;        matrixMul();        for(int i=1;i<=n;i++){            for(int j=1;j<=n-1;j++)                write((c[i][j]+3)%3),putchar(' ');            write((c[i][n]+3)%3),putchar('\n');        }    }    return 0;}

按照比赛的时候最早的想法：忽略0的值，1 * 1 = 1 ， 2 * 1 = 2 ，1 * 2 = 2 ，2 * 2 = 1

bitset记录上面，然后用 & 的 bitset 的count 计算组合后的个数

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define eps 1e-9#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=1000;int n,A,B;bitset <MAXN> a[3][MAXN],b[3][MAXN];void init(){for(int k=1;k<3;k++)for(int i=0;i<n;i++)  for(int j=0;j<n;j++){a[k][i].reset(j);b[k][j].reset(i);   }}int main(){while(read(n)){init();for(int i=0;i<n;i++)for(int j=0;j<n;j++){read(A);A%=3;a[A][i].set(j);}for(int i=0;i<n;i++)for(int j=0;j<n;j++){read(B);B%=3;b[B][j].set(i);}for(int i=0;i<n;i++)for(int j=0;j<n;j++){int ans=0;ans+=(a[1][i] & b[1][j]).count();ans+=(a[1][i] & b[2][j]).count()*2;ans+=(a[2][i] & b[1][j]).count()*2;ans+=(a[2][i] & b[2][j]).count();write(ans%3);if(j==n-1)puts("");elseputchar(' ');}}return 0;}

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