貪心...我都好奇自己當(dāng)時(shí)沒(méi)事干怎么就學(xué)了貪心...

當(dāng)然盡管這樣23道題我也沒(méi)寫(xiě)完...

正如gg所說(shuō)：我們要不厭其煩地寫(xiě)題解。（當(dāng)然不存在的

1455:An Easy Problem

As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form.

Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of '1's in whose binary form is the same as that in the binary form of I.

For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 '1's. The minimum integer, which is greater than "1001110" and also contains 4 '1's, is "1010011", i.e. "83", so you should output "83".

輸入One integer per line, which is I (1 <= I <= 1000000).

A line containing a number "0" terminates input, and this line need not be processed.

輸出One integer per line, which is J.

其實(shí)這題就是把一個(gè)數(shù)轉(zhuǎn)換為二進(jìn)制，1的數(shù)量要求一樣，要得到比給定數(shù)大的最小的數(shù)。

所以只要統(tǒng)計(jì)出有多少個(gè)一就好了。

不想解釋別的了qwqqqqq

#include<iostream> #include<cstdio> #include<algorithm> using namespace std; int gcd(int n) {int flag=0;while(n){if(n%2){flag++;}n/=2;}return flag; } int main() {int n;while(cin>>n){if(!n)break;int t=gcd(n);for(int i=n+1; ;i++){if(gcd(i)==t){cout<<i<<endl;break;}}}return 0; }

我們稱之為路的，無(wú)非是躊躇。

轉(zhuǎn)載于:https://www.cnblogs.com/Grigory/p/10159642.html

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