hdu5791Two

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=5791

题意：给定a,b两个数组，求有多少个公共子序列。

分析：设dp[i][j]表示a中前i个和b中前j个并且a[i]==b[j]匹配的子序列个数，转移的时候要用前缀和优化一下。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=1010;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000007;const int INF=1000000010;const ll MAX=1ll<<55;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;ll dp[N][N],sum1[N][N],sum2[N][N];int a[N],b[N];int main(){    int i,j,n,m;    ll ans;    while (scanf("%d%d", &n, &m)!=EOF) {        for (i=1;i<=n;i++) scanf("%d", &a[i]);        for (i=1;i<=m;i++) scanf("%d", &b[i]);        for (i=0;i<=n;i++)            for (j=0;j<=m;j++) dp[i][j]=sum1[i][j]=sum2[i][j]=0;        for (i=1;i<=n;i++)            for (j=1;j<=m;j++)            if (a[i]==b[j]) {                dp[i][j]=(sum2[i-1][j-1]+1)%MOD;                sum1[i][j]=(sum1[i-1][j]+dp[i][j])%MOD;                sum2[i][j]=(sum2[i][j-1]+sum1[i][j])%MOD;            } else sum1[i][j]=sum1[i-1][j],sum2[i][j]=(sum2[i][j-1]+sum1[i][j])%MOD;        ans=0;        for (i=1;i<=n;i++)            for (j=1;j<=m;j++) ans=(ans+dp[i][j])%MOD;        printf("%I64d\n", (ans+MOD)%MOD);    }    return 0;}

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