從左往右處理，左半部分記為left, 右半部分記為right,若i,i -1均為1， 貢獻為ans += (left + 1) + right * (1ll << (i - 1));

否則貢獻為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);}}

　　

轉載于:https://www.cnblogs.com/IMGavin/p/6024003.html

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