Kim Schrijvers
Consider an ordered set S of strings of N (1 <= N <= 31) bits. Bits, of course, are either 0 or 1.
This set of strings is interesting because it is ordered and contains all possible strings of length N that have L (1 <= L <= N) or fewer bits that are `1'.
Your task is to read a number I (1 <= I <= sizeof(S)) from the input and print the Ith element of the ordered set for N bits with no more than L bits that are `1'.
PROGRAM NAME: kimbits
INPUT FORMAT
A single line with three space separated integers: N, L, and I.
SAMPLE INPUT (file kimbits.in)
5 3 19
OUTPUT FORMAT
A single line containing the integer that represents the Ith element from the order set, as described.
SAMPLE OUTPUT (file kimbits.out)
10011
题解:看到这个果断枚举了。。。速度码出来了。。不过TLE了。。。N=31,枚举量等于2^31。。。线性的枚举也得超时啊。。。这样的话只能把第I个符合要求的数构造出来。
我们分两步:
第一步:计算长度为i,1的个数不超过j的01字符串组成的集合的个数,用f[i][j]表示。
那么DP方程为f[i][j]=f[i-1][j]+f[i-1][j-1](分别表示开头为‘0’和开头为‘1’的情况)。
边界条件:f[i,0]=1,f[0,j]=1。
第二步:这部就是利用上述DP方程构造出第I个符合要求的01字符串。估计我说得很渣。。。直接引用nocow上的话好了:设所求串为S,假设S的位中最高位的1在自右向左第K+1位,那么必然满足F[K,L]< I,F[K+1,L] >=I,这样的K是唯一的。所以S的第一个1在从右至左第K+1位.因为有F[K,L]个串第K+1位上为0,所以所求的第I个数的后K位就应该是满足"位数为K且串中1不超过L-1个"这个条件的第I-F[K,L]个数。
View Code
/* ID:spcjv51 PROG:kimbits LANG:C */ #include<stdio.h> #include<string.h> int main(void) { freopen("kimbits.in","r",stdin); freopen("kimbits.out","w",stdout); long i,j,n,l; long long m; long f[32][32]; scanf("%ld%ld%lld",&n,&l,&m); memset(f,0,sizeof(f)); for(i=0;i<=n;i++) { f[0][i]=1; f[i][0]=1; } for(i=1;i<=n;i++) for(j=1;j<=l;j++) f[i][j]=f[i-1][j]+f[i-1][j-1]; for(i=n;i>=1;i--) if(f[i-1][l]<m) { printf("1"); m-=f[i-1][l]; l--; } else printf("0"); printf("\n"); return 0; }