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原题连接：
Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, …, xm> another sequence Z = <z1, z2, …, zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, …, ik> of indices of X such that for all j = 1,2,…,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y. The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.问题描述 给定序列的子序列是给定序列，其中省略了一些元素（可能没有）。 给定序列X = <x1，x2，…，xm>另一个序列Z = <z1，z2，…，zk>是X的子序列，如果存在严格增加的序列<i1，i2，…, ik>对于所有j = 1,2，…，k，xij = zj，X的索引，…，ik>。 例如，Z = <a，b，f，c>是X = <a，b，c，f，b，c>的子序列，其索引序列<1,2,4,6>。 给定两个序列X和Y，问题是找到X和Y的最大长度公共子序列的长度。 程序输入来自文本文件。 文件中的每个数据集包含两个表示给定序列的字符串。 序列由任意数量的空格分隔。 输入数据是正确的。 对于每组数据，程序在标准输出上打印从单独行开始的最大长度公共子序列的长度。
Sample Input
abcfbc abfcab programming contest abcd mnpSample Output
4 2 0 个人思路：处理时可以选择字符的读取从第一个位置开始，或者把 i 号字符的状态存储到i+1号位置去，这样就从1号开始处理了。 ac代码：#include#include #include #define Max( a, b ) (a) > (b) ? (a) : (b)using namespace std;char s1[1005], s2[1005];int dp[1005][1005];int main(){ int len1, len2; while (scanf("%s %s", s1 + 1, s2 + 1) != EOF) { memset(dp, 0, sizeof(dp)); len1 = strlen(s1 + 1), len2 = strlen(s2 + 1); for (int i = 1; i <= len1; ++i) { for (int j = 1; j <= len2; ++j) { if (s1[i] == s2[j]) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = Max(dp[i - 1][j], dp[i][j - 1]); } } } printf("%d\n", dp[len1][len2]); } return 0;}
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