HDU 5791 Two (DP)

题意：给定两个串，让你求出，两个串字串，相同的个数。

析：dp[i][j] 表示 第一个第 i 个位置，第二串第 j 个位置，有多少相同的串，

如果  a[i] == b[j] 那么 dp[i][j] = dp[i-1][j-1] + dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + 1。

否则  dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1]

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r > 0 && r <= n && c > 0 && c <= m;
}

LL dp[maxn][maxn];
int a[maxn], b[maxn];

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
    for(int i = 1; i <= m; ++i)  scanf("%d", b+i);
    ms(dp, 0);
    for(int i = 1; i <= n; ++i)
      for(int j = 1; j <= m; ++j)
        if(a[i] == b[j])  dp[i][j] = (dp[i-1][j] + dp[i][j-1] + 1) % mod;
        else   dp[i][j] = (dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + mod) % mod;
    printf("%I64d
", dp[n][m]);
  }
  return 0;
}