8 solutions

  • 4
    @ 2021-10-4 16:57:07

    比如：

    (此图来自网络)

    代码如下：

    #include<cstdio>
#define ll long long

using namespace std;
ll a[505][505],b[505][505];
const ll mod=1000000007;
ll n,m,p;

inline ll read(){
	ll a=0,k=1;
	char c=getchar();
	while (!('0'<=c&&c<='9')){
		if (c=='-'){
			k*=-1;
		}
		c=getchar();
	}
	while ('0'<=c&&c<='9'){
		a=a*10+c-'0';
		c=getchar();
	}
	return a*k;
}

int main(){
	n=read(),p=read(),m=read();
	for (ll i=1;i<=n;i++){
		for (ll j=1;j<=p;j++){
			a[i][j]=read();
		}
	}
	for (ll i=1;i<=p;i++){
		for (ll j=1;j<=m;j++){
			b[i][j]=read();
		}
	}
	for (ll i=1;i<=n;i++){
		for (ll j=1;j<=m;j++){
			ll tmp=0;
			for (ll k=1;k<=p;k++){
				tmp=(tmp+(a[i][k]*b[k][j]+mod)%mod+mod)%mod;
			}
			printf("%lld ",tmp);
		}
		putchar('\n');
	}
	return 0;
}
    • 3
      @ 2022-6-9 15:19:14

      模板题，下面提供一份封装的矩乘板子：

      #include <cstring>
#include <iostream>

using std::cin;
using std::cout;
const char endl = '\n';

const int N = 1005,
          mod = 1e9 + 7;

int n, m, p;

class Matrix {
  private:
    int data[N][N];

  public:
    Matrix() {
        memset(data, 0x00, sizeof(data));
    }

    int* operator[](int i) {
        return data[i];
    }

    Matrix operator*(Matrix b) const {
        Matrix c;

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= p; j++) {
                for (int k = 1; k <= m; k++) {
                    c[i][j] = (c[i][j] + 1ll * data[i][k] * b[k][j] % mod) % mod;
                }
            }
        }

        return c;
    }
} a, b, c;

int main() {
    std::ios::sync_with_stdio(false);

    cin >> n >> m >> p;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> a[i][j];
        }
    }

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= p; j++) {
            cin >> b[i][j];
        }
    }

    c = a * b;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= p; j++) {
            cout << (c[i][j] % mod + mod) % mod << ' ';
        }
        cout << endl;
    }

    return 0;
}
      • 2
        @ 2023-1-25 14:01:17

        【模板】矩阵乘法

        这道题是一道模板题，还不会的同志们看下面这张图：

        image

        比如说： 给出一个矩阵A： 1 2 3 3 2 1 再给出矩阵B： 1 1 2 2 3 3 根据公式得出矩阵C： 14 14 10 10

        过程如下

        image

        那么按照公式模拟即可，别忘了取模！（尽量一边算一边mod 代码：

        #include <bits/stdc++.h>
using namespace std;
long long A[1114][1114],B[1114][1114],C[1114][1114];
const long long mod=1000000007;
int main(){
int n,m,k;
cin>>n>>m>>k;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>A[i][j];
}
}
for(int i=1;i<=m;i++){
for(int j=1;j<=k;j++){
cin>>B[i][j];
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=k;j++){
long long sum=0;
for(int l=1;l<=m;l++){
sum=(sum+(A[i][l]*B[l][j]+mod)%mod+mod)%mod;
}
C[i][j]=sum%1000000007;
//cout<<sum<<" ";
}
//cout<<endl;
}
//if(you use the grey one)return 0 here
for(int i=1;i<=n;i++){
for(int j=1;j<=k;j++){
cout<<C[i][j]<<" ";
}
cout<<endl;
}
return 0;

        完结撒花！！！

        • 1
          @ 2023-10-28 8:54:39 
          #include<bits/stdc++.h>
#define ll long long

using namespace std;
ll a[505][505],b[505][505];
const ll mod=1000000007;
ll n,m,p;

inline ll read(){**
	ll a=0,k=1;
	char c=getchar();
	while (!('0'<=c&&c<='9')){
		if (c=='-'){
			k*=-1;
		}
		c=getchar();
	}
	while ('0'<=c&&c<='9'){
		a=a*10+c-'0';
		c=getchar();
	}
	return a*k;
}

int main(){
	n=read(),p=read(),m=read();
	for (ll i=1;i<=n;i++){
		for (ll j=1;j<=p;j++){
			a[i][j]=read();
		}
	}
	for (ll i=1;i<=p;i++){
		for (ll j=1;j<=m;j++){
			b[i][j]=read();
		}
	}
	for (ll i=1;i<=n;i++){
		for (ll j=1;j<=m;j++){
			ll tmp=0;
			for (ll k=1;k<=p;k++){
				tmp=(tmp+(a[i][k]*b[k][j]+mod)%mod+mod)%mod;
			}
			printf("%lld ",tmp);
		}
		putchar('\n');
	}
	return 0;
}
          • 0
            @ 2023-3-25 9:37:55 
            #include<cstdio>
#define ll long long

using namespace std;
ll a[505][505],b[505][505];
const ll mod=1000000007;
ll n,m,p;

inline ll read(){
	ll a=0,k=1;
	char c=getchar();
	while (!('0'<=c&&c<='9')){
		if (c=='-'){
			k*=-1;
		}
		c=getchar();
	}
	while ('0'<=c&&c<='9'){
		a=a*10+c-'0';
		c=getchar();
	}
	return a*k;
}

int main(){
	n=read(),p=read(),m=read();
	for (ll i=1;i<=n;i++){
		for (ll j=1;j<=p;j++){
			a[i][j]=read();
		}
	}
	for (ll i=1;i<=p;i++){
		for (ll j=1;j<=m;j++){
			b[i][j]=read();
		}
	}
	for (ll i=1;i<=n;i++){
		for (ll j=1;j<=m;j++){
			ll tmp=0;
			for (ll k=1;k<=p;k++){
				tmp=(tmp+(a[i][k]*b[k][j]+mod)%mod+mod)%mod;
			}
			printf("%lld ",tmp);
		}
		putchar('\n');
	}
	return 0;
}
            • -5
              @ 2021-12-17 21:35:32 
              #include <bits/stdc++.h>
#include <windows.h>
#define ll long long
#define mod 1000000007
#define maxn 510
inline int read()
{
    int X = 0; bool f = false; char ch = getchar();
    while (ch > '9' || ch < '0') {f |= ch == '-'; ch = getchar();}
    while (ch <= '9' && ch >= '0') {X = (X << 3) + (X << 1) + (ch ^ 48); ch = getchar();}
    return f ? -X : X;
}
inline void write(ll X)
{
    if (X == 0) {putchar('0'); return;}
    if (X < 0) {/*putchar('-'); X = -X;*/ X += mod;}
    short cnt = 0; char ch[31];
    while (X) {ch[++cnt] = (X % 10) ^ 48; X /= 10;}
    while (cnt) putchar(ch[cnt--]);
    return;
}
int n, p, m;
ll A[maxn][maxn], B[maxn][maxn], ANS[maxn][maxn];
int main()
{
    n = read(); p = read(); m = read();
    for (int i = 1; i <= n; ++i) for (int j = 1; j <= p; ++j) A[i][j] = read();
    for (int i = 1; i <= p; ++i) for (int j = 1; j <= m; ++j) B[i][j] = read();
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= m; ++j)
        {
            for (int k = 1; k <= p; ++k)
            {
                ANS[i][j] = (ANS[i][j] + (A[i][k] * B[k][j] % mod)) % mod;
            }
            write(ANS[i][j]); putchar(' ');
        }
        putchar('\n');
    }
    system("pause");
    return 0;
}
              • -11
                @ 2021-9-26 8:13:50

                不会把还有人不知道矩阵乘法

                #include<bits/stdc++.h>
#define int long long//记得开longlong
using namespace std;
const int N = 600, M = 1e9 + 7;
int a[N][N], b[N][N];
inline int max(int x, int y){return x > y ? x : y;}
inline int min(int x, int y){return x > y ? y : x;}
inline int read(){
	int x = 0, f = 1; char c = getchar();
	while (!isdigit(c)){if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)){x = (x << 3) + (x << 1) + (c ^ 48); c = getchar(); }
	return x * f;
}
inline void write(int x){
	if (x < 0) {x = ~(x - 1); putchar('-');}
	if (x > 9) write(x / 10);
	putchar(x % 10 ^ 48);
}
signed main(){
	int n = read(), p = read(), m = read();
	for (int i = 1; i <= n; i ++)
	    for (int j = 1; j <= p; j ++)
	        a[i][j] = (read() + M) % M;//记得取模
	for (int i = 1; i <= p; i ++)
	    for (int j = 1; j <= m; j ++)
	        b[i][j] = (read() + M) % M;//记得取模
	for (int i = 1; i <= n; i ++){
	    for (int j = 1; j <= m; j ++){
	    	int sum = M;
	    	for (int k = 1; k <= p; k ++)
	    	    sum = (sum + a[i][k] * b[k][j]) % M;
	    	write(sum % M);putchar(' ');
		}
		puts("");
	}
	return 0;
}
                • -23
                  @ 2021-9-15 20:59:30

                  az，不会

                  • 1

                  【模板】矩阵乘法

                  Information

                  ID
                  60
                  Time
                  2000ms
                  Memory
                  256MiB
                  Difficulty
                  3
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