Pizza Serparation

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

using std::vector;

using std::queue;

using std::map;

using std::sort;

using std::string;

#define read(x) scanf("%d",&x)

#define reads(x) scanf("%s",x)

int cmp(const void * x, const void * y) {

#define datatype int

    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));

    //x < y

    return dx > dy ?  : -;

#undef datatype

}

int a[], nex[];

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    int n;

    read(n);

    for (int i = ; i < n; i++) {

        read(a[i]);

        nex[i] = i + ;

    }

    nex[n - ] = ;

    int ans = 0x3FFFFFFF;

    for (int i = ; i < n; i++) {

        int sum = ;

        for (int j = i;; j = nex[j]) {

            sum += a[j];

            if (sum >= ) break;

        }

        if ( * (sum - ) < ans) ans =  * (sum - );

    }

    printf("%d\n", ans);

    return ;

}

XK Segments

STL真是手残拯救者

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<string>

using std::vector;

using std::queue;

using std::map;

using std::sort;

using std::string;

using std::lower_bound;

using std::upper_bound;

#define read(x) scanf("%d", &x)

#define reads(x) scanf("%s", x)

#define write(x) printf("%d ", x)

#define writes(x) printf("%s ", x)

#define writeln(x) printf("%d\n", x)

#define writesln(x) printf("%s\n", x)

typedef long long llint;

/*data structure*/

/*data structure*/

int cmp(const void * x, const void * y) {

#define datatype int

    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));

    //x < y

    return dx > dy ?  : -;

#undef datatype

}

/*global variable*/

map<llint, int> mp;

vector<llint> v;

int sum[];

/*global variable*/

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    int n;

    llint a, ap, h, l, b, x, k;

    while (scanf("%d%lld%lld", &n, &x, &k) != EOF) {

        mp.clear();

        v.clear();

        for (int i = ; i < n; i++) {

            scanf("%lld", &a);

            if (mp.find(a) == mp.end()) v.push_back(a);

            mp[a]++;

        }

        sort(v.begin(), v.end());

        sum[] = mp[v[]];

        for (int i = ; i < v.size(); i++) sum[i] = sum[i - ] + mp[v[i]];

        llint ans = ;

        for (int i = ; i < v.size(); i++) {

            a = v[i];

            if (a % x == ) ap = a;

            else ap = a - a % x + x;

            if (k == ) {

                h = ap - ;

                l = a;

            } else {

                ap += x * (k - );

                l = ap;

                h = ap + x - ;

            }

            if (l > h) continue;

            int s = lower_bound(v.begin(), v.end(), l) - v.begin(), e = lower_bound(v.begin(), v.end(), h) - v.begin();

            if (e == v.end() - v.begin()) e--;

            if (v[e] > h) e--;

            if (s == v.end() - v.begin()) continue;

            ans += (llint)mp[a] * (llint)(sum[e] - sum[s] + mp[v[s]]);

        }

        printf("%lld\n", ans);

    }

    return ;

}

Square Subsets

首先，其中本来就是完全平方数的数，爱取多少取多少，不影响结果，只需最终结果乘以2^a[i]

对于不是完全平方数的数，只需关心其取了奇数个还是偶数个，具体数目不影响结果，因此可以看成只取0个或1个，最终结果乘以2^(a[i]-1)。

将1-70的数分解质因数，记录每个数的各个因子有奇数个还是偶数个。

1-70间的质数共19个，用0-2^19-1表示所有状态，第i为为1表示第i个质数有奇数个，为0表示有偶数个。

dp[i]表示达到i这种质因子奇偶状态的方法数，dp[0]表示完全平方数的方法数。

最终结果减1，去掉空集。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<string>

using std::vector;

using std::queue;

using std::map;

using std::sort;

using std::string;

using std::lower_bound;

using std::upper_bound;

#define read(x) scanf("%d", &x)

#define reads(x) scanf("%s", x)

#define write(x) printf("%d ", x)

#define writes(x) printf("%s ", x)

#define writeln(x) printf("%d\n", x)

#define writesln(x) printf("%s\n", x)

#define fillchar(x, a) memset(x, a, sizeof(x))

typedef long long llint;

/*data structure*/

/*data structure*/

int cmp(const void * x, const void * y) {

#define datatype int

    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));

    //x < y

    return dx > dy ?  : -;

#undef datatype

}

/*global variable*/

const int p[] = {, , , , , , , , , , , , , , , , , , };

const llint mod = ;

int a[], v[][];

llint dp[][ << ];

/*global variable*/

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    int n, x;

    read(n);

    fillchar(a, );

    fillchar(v, );

    for (int i = ; i < n; i++) {

        read(x);

        a[x]++;

    }

    int multp = ;

    for (int i = ; i <= ; i++) {

        multp += a[i * i];

        a[i * i] = ;

    }

    for (int i = ; i <= ; i++) {

        if (a[i]) multp += (a[i] - );

    }

    for (int i = ; i <= ; i++) {

        x = i;

        for (int j = ; j < ; j++) {

            while (x % p[j] == ) {

                v[i][j] =  - v[i][j];

                x /= p[j];

            }

        }

    }

    fillchar(dp, );

    dp[][] = ;

    int last = , cur;

    for (int i = ; i <= ; i++) {

        if (a[i] == ) continue;

        cur =  - last;

        for (int j = ; j < ( << ); j++) dp[cur][j] = ;

        for (int j = ; j < ( << ); j++) {

            int state = j;

            if (dp[last][state] == ) continue;

            for (int k = ; k < ; k++) {

                state ^= (v[i][k] << k);

            }

            dp[cur][state] = (dp[cur][state] + dp[last][j]) % mod;

        }

        for (int j = ; j < ( << ); j++) {

            dp[cur][j] = (dp[cur][j] + dp[last][j]) % mod;

        }

        last = cur;

    }

    llint ans = dp[last][];

    for (int i = ; i < multp; i++) {

        ans = ans *  % mod;

    }

    printf("%lld\n", ans - );

    return ;

}

String Mark

设f(s)为用字符串a组合能产生比s小的字符串的个数，答案为f(b)-f(a)-1，一位一位算就可以了。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<string>

#include<functional>

#include<iostream>

using std::priority_queue;

using std::vector;

using std::queue;

using std::map;

using std::sort;

using std::string;

using std::lower_bound;

using std::upper_bound;

using std::cin;

using std::cout;

using std::endl;

#define fillchar(x, a) memset(x, a, sizeof(x))

typedef long long lint;

/*data structure*/

/*data structure*/

int cmp(const void * x, const void * y) {

#define datatype int

    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));

    //x < y

    return dx > dy ?  : -;

#undef datatype

}

/*global variable*/

char a[], b[];

lint fac[], ifac[];

int cnt[], n;

const lint mod = ;

/*global variable*/

/*function*/

inline lint pow(lint a, lint b, lint p) {

    lint rtn = ;

    while (b) {

        if (b & ) rtn = rtn * a % p;

        a = a * a % p;

        b >>= ;

    }

    return rtn;

}

lint f(char * s) {

    fillchar(cnt, );

    for (int i = ; i < n; i++) cnt[a[i] - 'a']++;

    lint div = ;

    for (int i = ; i < ; i++) div = div * ifac[cnt[i]] % mod;

    lint rtn = ;

    for (int i = ; i < n; i++) {

        for (int j = ; j < s[i] - 'a'; j++) {

            if (cnt[j]) {

                rtn = (rtn + fac[n - i - ] * cnt[j] % mod * div % mod) % mod;

            }

        }

        if (cnt[s[i] - 'a'] == ) return rtn;

        div = div * cnt[s[i] - 'a'] % mod;

        cnt[s[i] - 'a']--;

    }

    return rtn;

}

/*function*/

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    std::ios::sync_with_stdio(), cin.tie();

    cin >> a >> b;

    n = strlen(a);

    fac[] = ifac[] = ;

    for (int i = ; i <= n; i++) {

        fac[i] = fac[i - ] * i % mod;

        ifac[i] = pow(fac[i], mod - , mod);

    }

    cout << (f(b) - f(a) -  + mod) % mod << endl;

    return ;

}

Eyes Closed

每个线段里随机选个数，选出来的数期望为线段里的数的平均数，设为a。对于长度为l的线段里的一个数字x，其保持不变的概率为(l-1)/l，被选出来交换的概率为1/l，则其期望为ex=x*(l-1)/l+a/l，即操作后线段里每个数字的期望为原数字乘以一个常数再加一个常数，可以用线段树来实现这个操作，把lazy标记分成两部分即可。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<string>

using std::vector;

using std::queue;

using std::map;

using std::sort;

using std::string;

using std::lower_bound;

using std::upper_bound;

#define read(x) scanf("%d", &x)

#define reads(x) scanf("%s", x)

#define write(x) printf("%d ", x)

#define writes(x) printf("%s ", x)

#define writeln(x) printf("%d\n", x)

#define writesln(x) printf("%s\n", x)

#define fillchar(x, a) memset(x, a, sizeof(x))

typedef long long llint;

/*data structure*/

template <class T> class SegmentTree {

public:

    T dat, lazym, lazyp;

    int leftBorder, rightBorder, mid;

    SegmentTree * leftSon, * rightSon;

    SegmentTree() {

        leftBorder = rightBorder = -;

        leftSon = rightSon = NULL;

    }

    void pushdown();

    void pushup();

    void Build(T *, int, int);

    void Modify(int, int, T, T);

    T Query(int, int);

    void Free();

};

template<class T> void SegmentTree<T>::pushdown() {

    if ((lazym !=  || lazyp != (T)) && leftBorder != rightBorder) {

        leftSon->dat = leftSon->dat * lazym + lazyp * (leftSon->rightBorder - leftSon->leftBorder + );

        rightSon->dat = rightSon->dat * lazym + lazyp * (rightSon->rightBorder - rightSon->leftBorder + );

        leftSon->lazym *= lazym;

        leftSon->lazyp = leftSon->lazyp * lazym + lazyp;

        rightSon->lazym *= lazym;

        rightSon->lazyp = rightSon->lazyp * lazym + lazyp;

    }

    lazym = (T);

    lazyp = (T);

}

template<class T> void SegmentTree<T>::pushup() {

    dat = leftSon->dat + rightSon->dat;

}

template<class T> void SegmentTree<T>::Build(T * S, int l, int r) {

    if (l > r) {

        return;

    }

    lazym = (T);

    lazyp = (T);

    leftBorder = l;

    rightBorder = r;

    mid = (leftBorder + rightBorder) >> ;

    if (l == r) {

        dat = S[l];

        return;

    }

    leftSon = new SegmentTree;

    leftSon->Build(S, l, mid);

    rightSon = new SegmentTree;

    rightSon->Build(S, mid + , r);

    pushup();

}

template<class T> void SegmentTree<T>::Modify(int l, int r, T mult, T plus) {

    if (l > r || l < leftBorder || rightBorder < r) {

        return;

    }

    if (leftBorder == l && rightBorder == r) {

        dat = dat * mult + plus * (rightBorder - leftBorder + );

        lazym *= mult;

        lazyp = lazyp * mult + plus;

        return;

    }

    pushdown();

    if (r <= mid) {

        leftSon->Modify(l, r, mult, plus);

    } else if (mid < l) {

        rightSon->Modify(l, r, mult, plus);

    } else {

        leftSon->Modify(l, mid, mult, plus);

        rightSon->Modify(mid + , r, mult, plus);

    }

    pushup();

}

template<class T> T SegmentTree<T>::Query(int l, int r) {

    if (l > r || l < leftBorder || rightBorder < r) {

        return dat;

    }

    pushdown();

    if (l == leftBorder && r == rightBorder) {

        return dat;

    }

    if (r <= mid) {

        return leftSon->Query(l, r);

    } else if (mid < l) {

        return rightSon->Query(l, r);

    } else {

        return leftSon->Query(l, mid) + rightSon->Query(mid + , r);

    }

}

template<class T> void SegmentTree<T>::Free() {

    if (leftSon != NULL) {

        leftSon->Free();

    }

    if (rightSon != NULL) {

        rightSon->Free();

    }

    delete leftSon;

    delete rightSon;

}

/*data structure*/

int cmp(const void * x, const void * y) {

#define datatype int

    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));

    //x < y

    return dx > dy ?  : -;

#undef datatype

}

/*global variable*/

SegmentTree<double> st;

double a[];

/*global variable*/

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    int n, q;

    read(n);

    read(q);

    for (int i = ; i < n; i++) scanf("%lf", &a[i]);

    st.Build(a, , n - );

    int t, l1, l2, l, r1, r2, r;

    for (int i = ; i < q; i++) {

        read(t);

        if (t == ) {

            read(l1);

            read(r1);

            read(l2);

            read(r2);

            l1--, r1--, l2--, r2--;

            double sum1 = st.Query(l1, r1), sum2 = st.Query(l2, r2);

            double ave1 = sum1 / (r1 - l1 + 1.0), ave2 = sum2 / (r2 - l2 + 1.0);

            st.Modify(l1, r1, (r1 - l1) / (r1 - l1 + 1.0), ave2 / (r1 - l1 + 1.0));

            st.Modify(l2, r2, (r2 - l2) / (r2 - l2 + 1.0), ave1 / (r2 - l2 + 1.0));

        } else {

            read(l);

            read(r);

            l--, r--;

            double ans = st.Query(l, r);

            printf("%.7lf\n", ans);

        }

    }

    return ;

}

Codeforces Round #448的更多相关文章

  1. Codeforces Round #448 C. Square Subsets

    题目链接 Codeforces Round #448 C. Square Subsets 题解 质因数 *质因数 ＝ 平方数,问题转化成求异或方程组解的个数 求出答案就是\(2^{自由元-1}\) , ...

  2. Codeforces Round #448(Div.2) Editorial ABC

    被B的0的情况从头卡到尾.导致没看C,心情炸裂又掉分了. A. Pizza Separation time limit per test 1 second memory limit per test ...

  3. Codeforces Round #448 (Div. 2) B

    题目描述有点小坑,ij其实是没有先后的 并且y并不一定存在于a中 判断y的个数和所给数组无关 对于2 - 7来说 中间满足%2==0的y一共有3个 2 4 6 这样 可以看出对于每个数字a 都能够二分 ...

  4. Codeforces Round #448 (Div. 2)C. Square Subsets

    可以用状压dp,也可以用线型基,但是状压dp没看台懂... 线型基的重要性质 性质一:最高位1的位置互不相同 性质二:任意一个可以用这些向量组合出的向量x,组合方式唯一 性质三:线性基的任意一个子集异 ...

  5. Codeforces Round #448 (Div. 2) B. XK Segments【二分搜索/排序/查找合法的数在哪些不同区间的区间数目】

    B. XK Segments time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  6. Codeforces Round #448 (Div. 2) A. Pizza Separation【前缀和/枚举/将圆（披萨）分为连续的两块使其差最小】

    A. Pizza Separation time limit per test 1 second memory limit per test 256 megabytes input standard ...

  7. Codeforces Round #366 (Div. 2) ABC

    Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...

  8. Codeforces Round #354 (Div. 2) ABCD

    Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/out ...

  9. Codeforces Round #368 (Div. 2)

    直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输 ...

随机推荐

  1. （转）OpenLayers3基础教程——OL3之Popup

    http://blog.csdn.net/gisshixisheng/article/details/46794813 概述: 本节重点讲述OpenLayers3中Popup的调用时实现,OL3改用O ...

  2. 【Android】进程间通信IPC——AIDL

    AIDL官网定义AIDL(Android 接口定义语言)与您可能使用过的其他 IDL 类似. 您可以利用它定义客户端与服务使用进程间通信 (IPC) 进行相互通信时都认可的编程接口. 在 Androi ...

  3. jQuery默认select选择第一个元素

    $("#id option:first").prop("selected", 'selected');

  4. 【剑指Offer】39、平衡二叉树

      题目描述:   输入一棵二叉树,判断该二叉树是否是平衡二叉树.这里的定义是:如果某二叉树中任意结点的左.右子树的深度相差不超过1,那么它就是一棵平衡二叉树.   解题思路:   首先对于本题我们要 ...

  5. 【数据结构】4.1图的创建及DFS深度遍历（不完善）

    声明:本代码仅供参考,根本就不是正确代码(至少在我看来,有很多BUG和不完美的地方) 图的存储方式选择为邻接表,并且headNode只是来存储一个链表的Node首地址额 总之这个代码写的很垃圾呀很垃圾 ...

  6. EXTJS之DATA PROXY READER

    这是不依赖于STORE的读取.我测试了很久,原来在新版本的EXTJS里.modelmanager.getmodel之类的不用了. 更改为静态的LOAD办法. <!DOCTYPE html> ...

  7. Leading and Trailing

    You are given two integers: n and k, your task is to find the most significant three digits, and lea ...

  8. Apache Shiro教程

    跟开涛学系列: 来自开涛的Apache Shiro教程:http://jinnianshilongnian.iteye.com/blog/2018398 附带的代码例子:https://github. ...

  9. Ext.TabPanel中的items具体解释

    Ext.TabPanel中的items: (来自项目源代码中的items条目代码) items:{ id:"opt1", title:"默认页面", tabTi ...

  10. formValidator向struts2 ajax传參验证

    话不多说,直接上代码 页面head里面的formvalidator验证配置 <script type="text/javascript"> $(document).re ...