topcoder-srm-595-div2

250分:
暴力就行了

/*************************************************************************    > File Name: 250.cpp    > Author: ALex    > Mail: [e-mail]     > 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: [e-mail]     > 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: [e-mail]     > 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);        }};