number 二分答案

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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);}}

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