uva10069 DistinctSubsequeces

dp[i][j]为a中的第i个与z中的d[j]各项匹配的结果没有写大数加法，后补上.#include <stdio.h>#include <string.h>int dp[10001][101]={0};int slove(char x[],char z[]){int m=strlen(z);int n=strlen(x);int count=0;int i,j,k;memset(dp,0,sizeof(dp));for(i=0;i<n;i++){for(j=0;j<m;j++){if(j==0){dp[i][j]=x[i]==z[j];}else{if(x[i]==z[j]){int k;for(k=0;k<i;k++){      dp[i][j]+=dp[k][j-1];}}}count+=dp[i][j]*(j==m-1);}}printf("%d\n",count);return 0;}int main(){int n;char x[10001],z[101];scanf("%d",&n);while(n){scanf("%s%s",x,z);slove(x,z);n--;}return 0;}