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
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把组合数求出来处理成前缀和
我们发现当I大于一段k长所有情况总和时，那么k+1位一定是1
那么这道题就做完了
其实最坑的一点，sizeof（S）=2147483648
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1 /* 2 ID: ivorysi 3 PROG: kimbits 4 LANG: C++ 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <set> 12 #include <vector> 13 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 14 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 15 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 16 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 17 #define inf 0x7fffffff 18 #define MAXN 400005 19 #define ivorysi 20 #define mo 97797977 21 #define ha 974711 22 #define ba 47 23 #define fi first 24 #define se second 25 //#define pis pair<int,string> 26 using namespace std; 27 typedef long long ll; 28 int c[40][40]; 29 int n,l; 30 ll k; 31 int num[40]; 32 void solve() { 33 scanf("%d%d%lld",&n,&l,&k); 34 siji(i,0,n) c[i][0]=1; 35 siji(i,1,n) { 36 int z=min(l,i); 37 siji(j,1,z) { 38 c[i][j]=c[i-1][j-1]+c[i-1][j]; 39 } 40 } 41 siji(i,0,n) siji(j,1,l) { 42 c[i][j]+=c[i][j-1]; 43 } 44 int cnt=l; 45 gongzi(i,n-1,0) { 46 if(c[i][cnt]<k) {k=k-c[i][cnt];num[i+1]=1;--cnt;} 47 } 48 gongzi(i,n,1) { 49 printf("%c",num[i]+'0'); 50 } 51 puts(""); 52 } 53 int main(int argc, char const *argv[]) 54 { 55 #ifdef ivorysi 56 freopen("kimbits.in","r",stdin); 57 freopen("kimbits.out","w",stdout); 58 #else 59 freopen("f1.in","r",stdin); 60 #endif 61 solve(); 62 }