http://oj.acm.zstu.edu.cn/JudgeOnline/problem.php?id=4233

題解：題目和解答完全不一樣，答案是101010101010101010......的字符串

題意是和二進制位有關的字符串排列。

某一位的狀態(tài)和這一位的序號-1的二進制中1個數(shù)的奇偶有關

C++版本一

AC版本

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp; ll l,r; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifscanf("%d",&t);while(t--){scanf("%lld%lld",&l,&r);if(l>r){swap(l,r);}temp=0;if(l%2==0){l++;}if(r%2==1){temp++;r--;}printf("%lld\n",(r-l+1)/2+temp);}//cout << "Hello world!" << endl;return 0; }

C++版本二

AC版本

#include<stdio.h> #include<algorithm> #include<string.h> #include<math.h> #include<vector> using namespace std; vector<int> V; int solve(int x) {if(x==0) return 0;if(x==1) return 1;int y=lower_bound(V.begin(),V.end(),x)-V.begin();if(V[y]==x) return x/2;int dif=V[y]-x;return solve(dif)+(x-dif)/2; } int main() {V.push_back(2);int x=2;while(x<=(int)1e9){x=x*2;V.push_back(x);}int T;scanf("%d",&T);while(T--){int l,r;scanf("%d%d",&l,&r);printf("%d\n",solve(r)-solve(l-1));}return 0; }

C++版本三

符合題意版本

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp; ll l,r; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//scanf("%d",&n);scanf("%d",&t);while(t--){scanf("%lld%lld",&l,&r);if(l>r){swap(l,r);}if(l%2==0){if(__builtin_popcount(l-1)%2==0)l--;elsel++;}if(r%2==1){if(__builtin_popcount(r-1)%2==0)r++;elser--;}printf("%lld\n",(r-l+1)/2);}//cout << "Hello world!" << endl;return 0; }

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