标签：mod main namespace space for #include using print ace
题意：
给出两个字符串，A,B，求A,B的lcs的数量。
题解：
在LCS的时候同时计算方案QAQ，可是傻逼的我不会，不会，不会。。。。。。
代码：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e3 + 7;
int dp[N][N], sum[N][N];
char ch1[N], ch2[N];
int main () {
scanf ("%s%s", ch1 + 1, ch2 + 1);
int n = strlen (ch1 + 1);
int m = strlen (ch2 + 1);
for (int i = 0; i <= n; ++i) sum[i][0] = 1;
for (int j = 0; j <= m; j++) sum[0][j] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (dp[i][j] < dp[i - 1][j]) {
dp[i][j] = dp[i - 1][j];
sum[i][j] = sum[i - 1][j];
}
else if (dp[i][j] == dp[i - 1][j]) {
sum[i][j] = (sum[i][j] + sum[i - 1][j]) % mod;
}
if (dp[i][j] < dp[i][j - 1]) {
dp[i][j] = dp[i][j - 1];
sum[i][j] = sum[i][j - 1];
}
else if (dp[i][j] == dp[i][j - 1]) {
sum[i][j] = (sum[i][j] + sum[i][j - 1]) % mod;
}
if (ch1[i] == ch2[j]) {
if (dp[i][j] < dp[i - 1][j - 1] + 1) {
dp[i][j] = dp[i - 1][j - 1] + 1;
sum[i][j] = sum[i - 1][j - 1];
}
else if (dp[i][j] == dp[i - 1][j - 1] + 1) {
sum[i][j] = (sum[i][j] + sum[i - 1][j - 1]) % mod;
}
}
}
}
printf("%d\n", sum[n][m]);
return 0;
}
总结：
一个基础的DP，我竟然不会QAQ，我真为“志几”感到捉急...
标签：mod main namespace space for #include using print ace
原文地址：http://www.cnblogs.com/xgtao/p/5987098.html
