topcoder-srm-595-div2

250分:
暴力就行了

/************************************************************************* > File Name: 250.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年05月15日 星期五 20时42分55秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

struct node {
    int l, r;
};

class LittleElephantAndBallsAgain {
    public:
        int getNumber(string str) {
            int n = str.length();
            int ans = inf;
            for (int i = 0; i < n;) {
                int l = i, r = i;
                while (i < n && str[i] == str[i + 1]) {
                    ++i;
                    ++r;
                }
                ans = min(ans, l - 1 + n - r);
                ++i;
            }
            return ans;
        }
};

500分:
状压然后hash下

/************************************************************************* > File Name: 500.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年05月15日 星期五 21时29分17秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

set <string> st;
char str[110];

class LittleElephantAndIntervalsDiv2 {
    public:
        int getNumber(int M, vector<int> l, vector<int> r) {
            int n = l.size();
            int ans = 0;
            st.clear();
            for (int i = 0; i < (1 << n); ++i) {
                for (int j = 0; j < M; ++j) {
                    str[j] = '0';
                }
                for (int j = 0; j < n; ++j) {
                    if (i & (1 << j)) {
                        for (int k = l[j] - 1; k <= r[j] - 1; ++k) {
                            str[k] = '1';
                        }
                    }
                    else {
                        for (int k = l[j] - 1; k <= r[j] - 1; ++k) {
                            str[k] = '0';
                        }
                    }
                }
                str[M] = '\0';
                if (st.find(str) == st.end()) {
                    st.insert(str);
                }
            }
            ans = st.size();
            return ans;
        }
};

1000分:
类似于数位dp
把A,B,C拆成二进制,然后用数位dp枚举某一位开始变小的方法去记忆化搜索一下
dp[cur][flag1][flag2][flag3] 表示到第cur位,A是否受到限制,B是否受到限制hi,C是否受到限制的方案数
原来想的是只拆A,B然后同样做法,最后如果得到的数字异或以后<=C,就返回1,不然就是0,但是不知道为什么每次算出来的都比答案大一些

/************************************************************************* > File Name: 1000.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年05月15日 星期五 22时12分33秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

LL dp[40][2][2][2];
int bitA[40], bitB[40], bitC[40];

LL dfs(int cur, bool flag1, bool flag2, bool flag3) {
    if (cur == -1) {
        return 1;
    }
    if (~dp[cur][flag1][flag2][flag3]) {
        return dp[cur][flag1][flag2][flag3];
    }
    LL ans = 0;
    int end1 = (flag1 ? bitA[cur] : 1);
    int end2 = (flag2 ? bitB[cur] : 1);
    for (int i = 0; i <= end1; ++i) {
        for (int j = 0; j <= end2; ++j) {
            int z = i ^ j;
            if (flag3 && z > bitC[cur]) {
                continue;
            }
            ans += dfs(cur - 1, flag1 && i == end1, flag2 && j == end2, flag3 && z == bitC[cur]);
        }
    }
    dp[cur][flag1][flag2][flag3] = ans;
    return ans;
}

LL calc(int A, int B, int C) {
    int cnt1 = 0, cnt2 = 0, cnt3 = 0;
    while (A) {
        bitA[cnt1++] = A % 2;
        A /= 2;
    }
    while (B) {
        bitB[cnt2++] = B % 2;
        B /= 2;
    }
    while (C) {
        bitC[cnt3++] = C % 2;
        C /= 2;
    }
    int maxs = max(cnt1, max(cnt2, cnt3));
    for (int i = cnt1; i < maxs; ++i) {
        bitA[i] = 0;
    }
    for (int i = cnt2; i < maxs; ++i) {
        bitB[i] = 0;
    }
    for (int i = cnt3; i < maxs; ++i) {
        bitC[i] = 0;
    }
    cnt1 = cnt2 = cnt3 = maxs;
    return dfs(cnt1 - 1, 1, 1, 1);
}

class LittleElephantAndXor {
    public:
        LL getNumber(int A, int B, int C) {
            memset(dp, -1, sizeof(dp));
            return calc(A, B, C);
        }
};