鏈接：https://ac.nowcoder.com/acm/contest/897/J
來源：牛客網

Binary Number
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
As a programmer, you are probably familiar with the binary representation of integers. That is, write an integer x as
∑
a
i
2
i
∑ai2i, where each
a
i
ai is either 0 or 1. Particularly, n-digit binary number can be written as
∑
n
?
1
i
=
0
a
i
2
i
∑i=0n?1ai2i, in which
a
n
?
1
an?1 must equal to 1.
This time, to test your mastery of binary numbers, Leg Han raises a problem to you.
Among all n-digit binary numbers whose amount of 1 is m, please print the k-th smallest one.
It is guaranteed that k is legal.
輸入描述:
The first line contains an integer number T, the number of test cases.

i
t
h
ith of each next T lines contains three integers n, m, k(
1
≤
m
≤
n
≤
31
,
1
≤
k
≤
2
×
10
8
1≤m≤n≤31,1≤k≤2×108).
輸出描述:
For each test case print the \(k\)-th smallest one.
示例1
輸入
復制
2
5 2 2
5 3 3
輸出
復制
10010
10110

題意：

思路：
我是反著來求的，要求第k小，我把所有情況sum算出來,令id=sum-k+1.然后求第id大的

從高位到低位，每一位我們考慮如果填了1，剩下還有種組合情況x，是否還夠達到第id大，如果夠就填1，不夠就填0，并且id-=x。這樣操作即可。

細節見代碼：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;} inline void getInt(int *p); const int maxn = 1000010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ll getpc(ll m, ll n) //m 中 選 n 個 {long long ans = 1;for (long long k = 1; k <= n; k++) {ans = (ans * (m - n + k)) / k;}return ans; } ll n, m, k; int main() {//freopen("D:\\code\\text\\input.txt","r",stdin);//freopen("D:\\code\\text\\output.txt","w",stdout);int t;gbtb;cin >> t;while (t--) {cin >> n >> m >> k;ll sum = getpc(n - 1, m - 1);ll id = sum - k + 1ll;// 反過來求第id大的數。ll temp = 0ll;ll rm = m;// 剩余1的數量for (ll i = n; i > 0; --i) {ll x = getpc(i - 1, rm - 1);if (x >= id && rm) {cout << 1;rm--;} else {id -= x;cout << 0;}cout << flush;}cout << endl;}return 0; }inline void getInt(int *p) {char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}} else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 + ch - '0';}} }

轉載于:https://www.cnblogs.com/qieqiemin/p/11218666.html

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