RoundNumber计数算法-优快云博客

// 160K 16MS    C++ 
#include <stdio.h>
#include <string.h>

int C[33][33]; // C[n][m], choose m from n;
int begin;
int end;

const int MAX = 0x80000000;

void getCombination() {
    for (int n = 0; n <= 32; n++) {
        for (int m = 0; m <= n; m++) {
            if (m == 0) { // choose nothing
                C[n][m] = 1;
            } else if (n == 0) { // nothing to choose
                C[n][m] = 0;
            } else if (m > n) {
                C[n][m] = 0;
            } else if (m == n) {
                C[n][m] = 1;
            } else {
                C[n][m] = C[n-1][m] + C[n-1][m-1];
                // printf("C %d %d is %d\n",n, m, C[n][m]);
            }
        }
    }
}

int getLength(int num) {
    int length = 0;
    while(num) {
        length++;
        num >>= 1;
    }
    return length;
}

int get1Num(int num) {
    int oneNum = 0;
    while(num) {
        if (num & 1) {
            oneNum++;
        }
        num >>= 1;
    }
    return oneNum;
}

int getRNnum(int num) {
    int length = getLength(num);
    int oneNum = get1Num(num);
    // printf("%x length: %d, 1Num: %d\n", num, length, oneNum);
    int sum = 0;

    //case 1: the most bit is 0:
    for (int i = 1; i < length-1; i++) {
        for (int j = i/2 + 1; j <= i; j++) { // +1 becasue most bit is 0
            sum += C[i][j];
            // printf("Step1+ %d %d %d\n", i, j, C[i][j]);
        }
    }
    // printf("S1 %d %d\n", num, sum);

    //case 2: the most bit is 1:
    int curNum = num;
    int curPos = 1;
    for (int i = 1; i <= oneNum-1; i++) {
        while(1) {
            if (curNum&1) {
                int waitFillPosNum = curPos - 1;
                int beforeCurPosLength = (length - curPos + 1);
                int oneNumBeforeCurPos = oneNum - i;
                int zeroNumBeforeCurPos = (length - curPos) - oneNumBeforeCurPos;

                int zeroMinNum;
                if (length%2) { // 1's num > 0's num
                    // zeroMinNum = (length - 2*zeroNumBeforeCurPos)/2;
                    zeroMinNum = length/2 - zeroNumBeforeCurPos;
                } else { // 1's num >= 0's num
                    // zeroMinNum = (length - 2*zeroNumBeforeCurPos)/2 - 1;
                    zeroMinNum = length/2 - zeroNumBeforeCurPos - 1;
                }
                // printf("s2 %d %d %d\n", waitFillPosNum, zeroNumBeforeCurPos, zeroMinNum);
                for (int j = zeroMinNum; j <= waitFillPosNum; j++) {
                    if (j >= 0) {
                        // printf("+ C[%d][%d]: %d\n", waitFillPosNum, j, C[waitFillPosNum][j]);
                        sum += C[waitFillPosNum][j];
                    }
                }
                curNum >>= 1;
                curPos++;
                break;
            } else {
                curPos++;
                curNum >>= 1;
            }
        }
    }
    // printf("%d %d %d\n", num, sum, num + 1 - sum);
    return sum;
}

int main() {
    getCombination();
    while(scanf("%d %d", &begin, &end) != EOF) {
        printf("%d\n", getRNnum(end + 1) - getRNnum(begin));
    }
}

http://blog.youkuaiyun.com/zhengnanlee/article/details/9794625, 一个很不错的解题报告.

http://www.cnblogs.com/Lyush/archive/2012/08/13/2636647.html DP解法:

组合数学一直没有系统的学过，刷的吃力，并且，这种题很考察细节，要专门做一批这样的题，提升自己处理细节的能力。

题目本质上用的就是简单的组合数学中的加法原理，只不过在一开始建组合表的时候要用到组合的一些定理.上面的报告已经将的很清晰了，就不赘述了.

思路就是先求出从0到某个数的roundNumber的个数，简称 RN()，然后对于从a到b之前的roundNumber的个数，直接 RN(b+1) - RN(a) 就是。

要求小于等于某个数的RN()数量，分为两种情况：

举个例子来说明：

2100 二进制是 1111101000

<1>: 0XXXXXXXXX(X代表0/1)，因为最高位是0，所以必定小于2100,

这种情况下，有以下的情况：

01XXXXXXXX

001XXXXXXX