hdu4055 dp
http://acm.hdu.edu.cn/showproblem.php?pid=4055
Your task is as follows: You are given a string describing the signature of many possible permutations, find out how many permutations satisfy this signature.
Note: For any positive integer n, a permutation of n elements is a sequence of length n that contains each of the integers 1 through n exactly once.
Each test case occupies exactly one single line, without leading or trailing spaces.
Proceed to the end of file. The '?' in these strings can be either 'I' or 'D'.
II ID DI DD ?D ??
1 2 2 1 3 6HintPermutation {1,2,3} has signature "II". Permutations {1,3,2} and {2,3,1} have signature "ID". Permutations {3,1,2} and {2,1,3} have signature "DI". Permutation {3,2,1} has signature "DD". "?D" can be either "ID" or "DD". "??" gives all possible permutations of length 3.
/** hdu4055 dp 题目大意:给定一个字符串,I表示本字符要比前一个字符大,D表示本字符要不前一个字符小,?可大可小。问1~n的全部排列中, 有多少满足条件 解题思路:能够用dp[i][j]表示:处理完第i位,序列末尾位j的序列共同拥有多少个。最后的结果为sigma{dp[N][i]},1≤i≤N 处理dp[1~i][]的过程中i是依次1~n相加。处理完dp[i-1][]后,增加的数即为i,而dp[i][j]是要将i放进去j换 出来,而这里有一种将i放进去j换出来。同一时候不影响升降顺序的方法是: 将dp[i-1][j]的i-1个数的序列中 ≥j 的数都加1。这样i-1变成了i,j变成了j+1,而j自然就补在后面了。所以对”ID“序列依次处理就可以,初始条件:dp[1][1] = 1; 即仅仅有{1}。 处理‘I’:dp[i][j] = sigma{dp[i-1][x]},当中1≤x≤j-1,可进一步简化,dp[i][j] = dp[i][j-1]+dp[i-1][j-1] 处理‘D’:dp[i][j] = sigma{dp[i-1][x]},当中j≤x≤i-1。可进一步简化。dp[i][j] = dp[i-1][j+1]+dp[i-1][j] 处理‘?
’:dp[i][j] = sigma{dp[i-1][x]}。当中1≤x≤i-1 */ #include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> using namespace std; const int mod=1e9+7; char a[1005]; int dp[1005][1005]; int main() { while(~scanf("%s",a)) { int n=strlen(a)+1; memset(dp,0,sizeof(dp)); dp[1][1]=1; for(int i=2;i<=n;i++) { char ch=a[i-2]; if(ch=='?') { int sum=0; for(int j=1;j<i;j++) { sum=(sum+dp[i-1][j])%mod; } for(int j=1;j<=i;j++) { dp[i][j]=sum; } } else if(ch=='I') { for(int j=2;j<=i;j++) { dp[i][j]=(dp[i][j-1]+dp[i-1][j-1])%mod; } } else { for(int j=i-1;j>=1;j--) { dp[i][j]=(dp[i][j+1]+dp[i-1][j])%mod; } } } int ans=0; for(int i=1;i<=n;i++) { ans=(ans+dp[n][i])%mod; } printf("%d\n",ans); } return 0; }