UVA 11645 Bits(组合数学)

从左往右处理，左半部分记为left, 右半部分记为right,若i,i -1均为1， 贡献为ans += (left + 1) + right * (1ll << (i - 1));

否则贡献为ans += right * (1ll << (i - 1));

import java.math.BigInteger;
import java.util.Scanner;
public class Main{
    public static void main(String[] args){
        Scanner cin = new Scanner(System.in);
        int t = 0;
        while(cin.hasNext()){
            BigInteger n = cin.nextBigInteger();

            if(n.compareTo(BigInteger.ZERO) < 0){
                break;
            }
            long cnt = n.toString(2).length();
            BigInteger right = n.shiftRight(2), left = BigInteger.ZERO;

            BigInteger ans = BigInteger.ZERO;
            for(long i = 1; i < cnt; i++){
            if(n.testBit((int)i) && n.testBit((int)i - 1)){
                ans = ans.add(left.add(BigInteger.ONE)).add(right.multiply(BigInteger.ONE.shiftLeft((int)i - 1)));
                //ans += (left + 1) + right * (1ll << (i - 1));
            }else{
                ans = ans.add(right.multiply(BigInteger.ONE.shiftLeft((int)i - 1)));
                //ans += right * (1ll << (i - 1));
            }
            right = right.shiftRight(1);
            if(n.testBit((int)i - 1)){
                left = left.add(BigInteger.ONE.shiftLeft((int)i - 1));
            }
            //left |= n & (1ll << (i - 1));
        }
        System.out.printf("Case %d: ", ++t);
        System.out.println(ans);

        }
        System.exit(0);

    }

}