HDU4055 Number String(카운트 dp)

Number String

전송문 1 전송문 2 The signature of a permutation is a string that is computed as follows: for each pair of consecutive elements of the permutation, write down the letter'I'(increasing) if the second element is greater than the first one, otherwise write down the letter'D'(decreasing).For example, the signature of the permutation {3,1,2,7,4,6,5} is “DIIDID”.
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.
Input
Each test case consists of a string of 1 to 1000 characters long, containing only the letters ‘I’, ‘D’ or ‘?’, representing a permutation signature.
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’.
Output
For each test case, print the number of permutations satisfying the signature on a single line. In case the result is too large, print the remainder modulo 1000000007.
Sample Input
II ID DI DD ?D ??
Sample Output
1 2 2 1 3 6
Hint
Permutation {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.

제목의 뜻

1-n의 배열을 정하면 I는 이전보다 크고, D는 이전보다 작다,?크든 작든 몇 가지 배열이 조건을 충족시키느냐고 물었다.

분석하다.

정의 dp[i][j]는 전 i위 서열의 끝이 j인 서열이 모두 몇 개인지 나타낸다.입력할 때scanf("%s",A+2);를 사용하는 것이 편리합니다.그러면 이 문자가 이전보다 크면 (A[i]=='I':
dp[i][j]=dp[i][j−1]+dp[i−1][j−1]
해당A[i]=='D'시:
dp[i][j]=dp[i][j+1]+dp[i−1][j+1]
해당A[i]=='?'시:
dp[i][j]=∑x=1i−1dp[i−1][x]
분명히 경계는
dp[1][1]=1.
시간 복잡도
O(n2T) (
T는 데이터 그룹 수)
참고 자료

CODE

#include
#include
#include
#define mod 1000000007
#define N 1005
#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)
#define ROF(i,a,b) for(int i=(a),i##_END_=(b);i>=i##_END_;i--)
char a[N];
int dp[N][N];
//dp[i][j]      i ，    j        。
inline int Add(int x,int y) {
    int t=x+y;
    if(t>=mod)t-=mod;
    return t;
}

int main() {
    while(~scanf("%s",a+2)) {
        int n=strlen(a+2)+1;
        memset(dp,0,sizeof(dp));
        dp[1][1]=1;
        FOR(i,2,n) {
            if(a[i]=='D')ROF(j,i-1,1)
                dp[i][j]=Add(dp[i][j+1],dp[i-1][j]);
            else if(a[i]=='I')FOR(j,2,i)
                dp[i][j]=Add(dp[i][j-1],dp[i-1][j-1]);
            else {
                int sum=0;
                FOR(j,1,i-1)sum=Add(sum,dp[i-1][j]);
                FOR(j,1,i)dp[i][j]=sum;
            }
        }
        int ans=0;
        FOR(i,1,n)ans=Add(ans,dp[n][i]);
        printf("%d
",ans);
    }
    return 0;
}

물론 각 층의 dp가 상층에서 이동하기 때문에 공간을 최적화할 수 있다.이 시간에 왜 몇 문제 안 써요?

#include
#include
#include
#define mod 1000000007
#define N 1005
#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)
#define ROF(i,a,b) for(int i=(a),i##_END_=(b);i>=i##_END_;i--)
char a[N];
int dp[2][N];
inline int Add(int x,int y) {
    int t=x+y;
    if(t>=mod)t-=mod;
    return t;
}

int main() {
    while(~scanf("%s",a+2)) {
        int n=strlen(a+2)+1;
        memset(dp,0,sizeof(dp));
        dp[1][1]=1;
        FOR(i,2,n) {
            int cur=i&1;
            if(a[i]=='D') {
                dp[cur][i]=0;
                ROF(j,i-1,1)
                    dp[cur][j]=Add(dp[cur][j+1],dp[!cur][j]);
            } else if(a[i]=='I') {
                dp[cur][1]=0;
                FOR(j,2,i)
                    dp[cur][j]=Add(dp[cur][j-1],dp[!cur][j-1]);
            } else {
                int sum=0;
                FOR(j,1,i-1)sum=Add(sum,dp[!cur][j]);
                FOR(j,1,i)dp[cur][j]=sum;
            }
        }
        int ans=0;
        FOR(i,1,n)ans=Add(ans,dp[n&1][i]);
        printf("%d
",ans);
    }
    return 0;
}