number 二分答案

题意

给出m和k. 某数i满足, i + 1, i + 2, i + 3.... 2 * i中恰好有m个数二进制拆分后有k位为1. 求出longlong范围以内， 输出一个这样的数和这样的数有多少个. 若有无穷个则输出-1.

题解

设f[i][j]表示第i个数中i + 1 ... 2 * i中恰好有j个1的个数. 不难发现， j一定时， 随i的单增， 值也呈单调性变化. 所以我们二分答案即可. 因为值单调， 所以一定只有一段连续区间是符合条件的.   所以二分上界和下界就能知道个数. 计算f[i][j]即可. 用比2*i小的二进制为k的个数-比i小的二进制为k的个数， 就是i- 2*i之间的.

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long dnt;
const int maxbit = 63;
const dnt inf = (dnt) 1 << 62;
int T, k, top, cnt, s[64];
dnt ans, ans1, ans2, m, c[64][64];
inline void init(){
	for(int i = 0; i <= maxbit; ++i) c[i][0] = 1;
	for(int i = 1; i <= maxbit; ++i)
		for(int j = 0; j <= maxbit; ++j)
			c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}
dnt dfs(dnt x,int cnt,int k){
	if(!k)return 1;
	if(cnt == -1) return 0;
	int p = ((x >> cnt) & 1);
	if(p) return dfs(x, cnt - 1, k - 1) + c[cnt][k];
	else  return dfs(x, cnt - 1, k);
}
inline dnt check(dnt x){
	return dfs(x * 2, maxbit, k) - dfs(x, maxbit, k);
}
int main(){
	freopen("number.in", "r", stdin);
	freopen("number.out", "w", stdout);
	init();
	scanf("%d", &T);
	while(T--){
		scanf("%I64d%d", &m, &k);
		if(m == 0){
			printf("%I64d %I64d\n", ((dnt) 1 << (k - 1)) - 1, ((dnt) 1 << (k - 1)) - 1);
			continue;
		}
		if(k == 1){
			printf("4 -1\n");
			continue;
		}
		dnt lf = 0, rg = inf, ans1 = 0, ans2 = 0;
		while(lf <= rg){
			dnt mid = (lf + rg) >> 1;
			if(check(mid) >= m) ans1 = mid, rg = mid - 1;
			else lf = mid + 1;
		}
		lf = 1, rg = inf;
		while(lf <= rg){
			dnt mid = (lf + rg) >> 1;
			if(check(mid) <= m) ans2 = mid, lf = mid + 1;
			else rg = mid - 1;
		}
		printf("%I64d %I64d\n", ans1, ans2 - ans1 + 1);
	}
}