Problem C. String Game
Input file: standard input
Output file: standard output
Time limit: 1 second
Memory limit: 256 megabytes
Clair and Bob play a game. Clair writes a string of lowercase characters, in which Bob sets the puzzle
by selecting one of his favorite subsequence as the key word of the game. But Bob was so stupid that he
might get some letters wrong.
Now Clair wants to know whether Bob’s string is a subsequence of Clair’s string and how many times
does Bob’s string appear as a subsequence in Clair’s string. As the answer may be very large, you should
output the answer modulo 109 + 7.
Input
The first line is Clair’s string (whose length is no more than 5000), and the second line is Bob’s string
(whose length is no more than 1000).
Output
Output one line, including a single integer representing how many times Bob’s string appears as a
subsequence in Clair’s string. The answer should modulo 109 + 7.
Examples
standard input standard output
eeettt
et
eeettt
te
9
0
题意：
给你一串主串和一串子串，让你在主串中匹配子串，匹配子串不需要是连续的子串，只要和子串字符出现顺序一样即可算作成功匹配。
题解：
dp，写出转移方程递推即可
AC代码：
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<iomanip>
#include<map>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<set>
#include<cctype>
#include<string>
#include<stdexcept>
#include<fstream>
#define mem(a,b) memset(a,b,sizeof(a))
#define dedebug() puts("what the fuck!")
#define debug(a) cout<<#a<<"="<<a<<endl;
#define speed {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); };
#define ll long long
using namespace std;
const double PI = acos(-1.0);
const int maxn = 1e6 + 50;
const int INF = 0x3f3f3f3f;
const double esp_0 = 1e-6;
const int mod = 1e9 + 7;
ll gcd(ll x, ll y) {
return y ? gcd(y, x % y) : x;
}
ll lcm(ll x, ll y) {
return x * y / gcd(x, y);
}
ll extends_gcd(ll a, ll b, ll& x, ll& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
ll gcdd = extends_gcd(b, a % b, x, y);
ll temp = x;
x = y;
y = temp - (a / b) * y;
return gcdd;
}
char str[5010];
char sub[1010];
int dp[1010][5010];
int main() {
while (~scanf("%s%s", str + 1, sub + 1)) {
mem(dp, 0);
int lenstr = strlen(str + 1);
int lensub = strlen(sub + 1);
for (int i = 1; i <= lensub; ++i) {
for (int j = 1; j <= lenstr; ++j) {
if (sub[i] == str[j]) {
if (i == 1) {
dp[i][j] = (dp[i][j - 1] + 1) % mod;
}
else {
dp[i][j] = (dp[i - 1][j - 1] + dp[i][j - 1]) % mod;
}
}
else {
dp[i][j] = dp[i][j - 1];
}
}
}
printf("%d\n", dp[lensub][lenstr]);
}
return 0;
}