Uva-10069 Distinct Subsequences DP

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题目链接

题目大意：给你两个字符串Str，str,求Str的str子序列种数。

状态： dp[i][j] 表示前Str的前str的子序列种树。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<math.h>#include<functional>#include<algorithm>#include<vector>#include<queue>using namespace std;const int maxn = 155;const int inf = 1<<29;const int mod = 100000000;int n;char Str[maxn*maxn],str[maxn];struct node{int len,num[15];}dp[3][maxn],One;void add( node &a,node &b) //大数相加{int c = 0;a.len = a.len > b.len?a.len:b.len;for( int i = 0; i < a.len; i ++ ){a.num[i] += b.num[i] + c;c = a.num[i]/mod;a.num[i] %= mod;}if( c )a.num[a.len++] = c;}void GetDp()      //核心代码{int Len = strlen(Str);int len = strlen(str);memset( dp,0,sizeof(dp) );for( int i = 1; i <= Len; i ++ ){for( int j = 1; j <= len; j ++ ){dp[i%2][j] = dp[(i-1)%2][j];if( Str[i-1] == str[j-1] ){if( j == 1 )add( dp[i%2][j],One );add( dp[i%2][j],dp[(i-1)%2][j-1] );}}}printf("%d",dp[Len%2][len].num[dp[Len%2][len].len-1]);for( int i = dp[Len%2][len].len-2; i >= 0; i -- ){printf("%08d",dp[Len%2][len].num[i]);}puts("");}int main() {#ifndef ONLINE_JUDGE    freopen("data.txt","r",stdin);#endifint cas;One.len = 1; One.num[0] = 1;scanf("%d ",&cas);while( cas -- ){scanf("%s %s",Str,str);GetDp();}    return 0;}

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