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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‘.
A single line with three space separated integers: N, L, and I.
5 3 19
A single line containing the integer that represents the Ith element from the order set, as described.
10011
题意：考虑排好序的N(N<=31)位二进制数。他们是排列好的，而且包含所有长度为N且这个二进制数中1的位数的个数小于等于L(L<=N)的数。你的任务是输出第i（1<=i<=长度为N的二进制数的个数）小的（注：题目这里表述不清，实际是，从最小的往大的数，数到第i个符合条件的，这个意思），长度为N，且1的位数的个数小于等于L的那个二进制数。（例：100101中，N=6，含有位数为1的个数为3）。
此题 I 最大值会爆int，= =
dp[i][j] 表示 i位的含j个1的二进制数个数，dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
/* ID: LinKArftc PROG: kimbits LANG: C++ */ #include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <vector> #include <cstdio> #include <string> #include <utility> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define randin srand((unsigned int)time(NULL)) #define input freopen("input.txt","r",stdin) #define debug(s) cout << "s = " << s << endl; #define outstars cout << "*************" << endl; const double PI = acos(-1.0); const int inf = 0x3f3f3f3f; const int INF = 0x7fffffff; typedef long long ll; const int maxn = 35; ll dp[maxn][maxn]; //dp[i][j]: i位的含j个1的二进制数个数 int n, l; ll k; void init() { dp[0][0] = dp[1][0] = dp[1][1] = 1; for (int i = 1; i < maxn; i ++) { dp[i][0] = 1; for (int j = 1; j < maxn; j ++) dp[i][j] = dp[i-1][j] + dp[i-1][j-1]; } } int main() { freopen("kimbits.in", "r", stdin); freopen("kimbits.out", "w", stdout); init(); scanf("%d %d %lld", &n, &l, &k); for (int i = 1; i <= n; i ++) { if (l == 0) { printf("0"); continue; } ll sum = 0; for (int j = 0; j <= l; j ++) { sum += dp[n-i][j]; } if (k <= sum) printf("0"); else { printf("1"); l --; k -= sum; } } printf("\n"); return 0; }
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原文地址：http://www.cnblogs.com/LinKArftc/p/5005999.html
