Codeforces 844B - Rectangles - 思维

链接:

  http://codeforces.com/contest/844/problem/B

题目：

You are given n × m table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:

All cells in a set have the same color.

Every two cells in a set share row or column.

Input

The first line of input contains integers n and m (1 ≤ n, m ≤ 50) — the number of rows and the number of columns correspondingly.

The next n lines of input contain descriptions of rows. There are m integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.

Output

Output single integer — the number of non-empty sets from the problem description.

Examples

input

1 1

0

output

1

input

2 3

1 0 1

0 1 0

output

8

Note

In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.

题意：

  给你一个n×m的棋盘，上面有0、1两种棋子。同种且同行（或同列）的棋子可以形成一个集合，一个棋子也可以看做一个集合。问你最多可以有多少种集合。

思路：

  考虑到n和m都比较小，分别枚举每一行的0、1棋子，对于每个棋子有选和不选两种可能。比如说某一行有k个1棋子，那么这一行可以形成的集合的数目就是2k−1，减一是因为要减去一个棋子都不选的这种不合法状态。枚举完行之后列同理。

  枚举完行和列之后会发现每个集合只有一个一个棋子的情况都被枚举了两遍，所以最终答案再减去n×m即可。

实现：

#include <iostream>
#include <algorithm>
#include <set>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <iomanip>
#include <functional>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#define edl putchar('\n')
#define ll long long
#define clr(a,b) memset(a,b,sizeof a)
#define rep(i,m,n) for(int i=m ; i<=n ; i++)
inline int read_() {int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9') {if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; }
namespace FastIO {
const int SIZE = 1 << 16;
char buf[SIZE], obuf[SIZE], str[60];
int bi = SIZE, bn = SIZE, opt;
int read(char *s) {
while (bn) {for (; bi < bn && buf[bi] <= ' '; bi++);if (bi < bn) break;
bn = fread(buf, 1, SIZE, stdin);bi = 0;}int sn = 0;while (bn) {
for (; bi < bn && buf[bi] > ' '; bi++) s[sn++] = buf[bi];if (bi < bn) break;
bn = fread(buf, 1, SIZE, stdin);bi = 0;}s[sn] = 0;return sn;}
bool read(int& x) {int n = read(str), bf;if (!n) return 0;
int i = 0; if (str[i] == '-') bf = -1, i++; else bf = 1;
for (x = 0; i < n; i++) x = x * 10 + str[i] - '0';if (bf < 0) x = -x;return 1;}};
#define read(x) FastIO::read(x)
#define read2(x,y) read(x),read(y)
#define read3(x,y,z) read(x),read(y),read(z)
#define readin read_()

using namespace std;

const int maxn = 57, mod = int(1e9)+7;
int mp[maxn][maxn];

int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
printf("init! init! init!\n");
#endif
int n, m, cnt1, cnt0; read2(n, m);
ll ans = 0;
rep(i,1,n) rep(j,1,m) read(mp[i][j]);
rep(i,1,n) {
cnt0 = cnt1 = 0;
rep(j,1,m) mp[i][j] ? cnt1++ : cnt0++;
ans += (1ll<<cnt0)-1, ans += (1ll<<cnt1)-1;
}
rep(i,1,m) {
cnt0 = cnt1 = 0;
rep(j,1,n) mp[j][i] ? cnt1++ : cnt0++;
ans += (1ll<<cnt0)-1, ans += (1ll<<cnt1)-1;
}
printf("%lld\n", ans-n*m);
return 0;
}