Divide Sum Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description long long ans = 0;for(int i = 1; i <= n; i ++)    for(int j = 1; j <= n; j ++)        ans += a[i] / a[j];给出n,a…

Sum Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description You are given an N*N digit matrix and you can get several horizontal or vertical digit strings from any position. For…

Sum vs Product Time Limit: 4000/2000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description Peter has just learned mathematics. He learned how to add, and how to multiply. The fact that 2 + 2 = 2 × 2…

Sum vs Product Time Limit: 2000/1000MS (Java/Others)    Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Peter has just learned mathematics. He learned how to add, and how to multiply. The fact that 2 + 2 = 2 × 2 has amaze…

先上题目: Power Sum Time Limit: 20000/10000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出n,m,p,求 (1^m + 2^m + 3^m + 4^m + ... + n^m) % p Input 第一行一个数T( <= 10),表示数据总数 然后每行给出3个数n,m,p(1 <= n <= m <= 10…

先上题目: C - Lowbit Sum Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description long long ans = 0;for(int i = 1; i <= n; i ++)    ans += lowbit(i)lowbit(i)的意思是将i转化成二进制数之后,只保留最低位的1及其后面的0,截断前面的内容,然…

J - Sum Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description You are given an N*N digit matrix and you can get several horizontal or vertical digit strings from any position. For example: 1…

Codeforces 963 A. Alternating Sum 题目大意:给出一组长度为n+1且元素为1或者-1的数组S(0~n),数组每k个元素为一周期,保证n+1可以被k整除.给a和b,求对1e9+9取模的结果 思路:容易想到,每个周期的∑组成的数列成等比,公比q=(b/a)^k,因此可以用等比数列公式求和.为了保证时间复杂度,需要用到快速幂运算:为了防止中间过程值溢出,需要多处取模,其中用费马小定理求逆元: 代码: #include<iostream> #include<cst…

J-Super Sum 题目大意就是给定N个三元组<a,b,c>求Σ(a1^k1*a2^k2*...*ai^ki*..an^kn)(bi<=ki<=ci) 唉.其实题目本身不难的,怪我不知道当时怎么想的...本来观察式子很容易能得出结论: 比如<5,2,3>,<2,1,4>,<3,2,2>这组: 5^2*2^1*3^2+5^2*2^2*3^2+5^2*2^3*3^2+5^2*2^4*^32=5^2*(2^1+2^2+2^3+2^4)*3^2; 5…

题意:有一个长度为\(2n\)数组,从中选分别选\(n\)个元素出来组成两个序列\(p\)和\(q\),(\(p\)和\(q\)中只要有任意一个元素在\(a\)的原位置不同,就算一个新的情况),选完后对\(p\)非降序排序,对\(q\)非升序排序,然后求它们每个元素对应位置的差的绝对值之和\(re s=\sum^{n}_1 |x_i-y_i|\),问所有情况的res总和. 题解:观察样例,不难发现,因为\(p\)非降序,\(q\)非升序,所以无论\(p\)和\(q\)怎么选,它们的贡献永远是排序…

GCD SUM Time Limit: 8000/4000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description 给出N,M执行如下程序:long long  ans = 0,ansx = 0,ansy = 0;for(int i = 1; i <= N; i ++)   for(int j = 1; j <= M; j ++)     …

先贴代码,以后再写题解... 首先,直接枚举肯定是会超时的,毕竟n就有10^9那么多... 对于每个数,我们先把它转化为二进制:例:21-->10101: 对于00001~10101,可以分为几个部分: 00001~10000: 10001~10100: 10101 因为对于每个数,从最右边的1截断,于是就可以理解为为: 00001~10000:: 001~100: 1: 设s[i]为二进制从右边数第 i+1 个数为1 (且其他数都为0)的lowbit sum: 则 s[i]=s[i-1]*2+…

题目链接:http://codeforces.com/problemset/problem/963/A 题目大意:就是给了你n,a,b和一段长度为k的只有'+'和‘-’字符串,保证n+1被k整除,让你你计算. 解题思路: 暴力肯定超时的,我们可以先计算出0~k-1这一段的值,当做a1,可以发现如果把每段长度为k的段的值当做一个元素,他们之间是成等比的,比值q=(b/a)^k, 然后就直接用等比数列求和公式求出答案即可.昨天把q当成b/a了,我的脑子啊... 注意,判断q==1时不能通过判断a==…

先上题目: Sum Time Limit: 6000/3000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出N,a[1]... a[N],还有M,b[1]... b[M]long long ans = 0;for(int i = 1; i <= N; i ++)    for(int j = 1; j <= M; j ++)        ans +=…

Lowbit Sum Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description long long ans = 0; for(int i = 1; i <= n; i ++)     ans += lowbit(i) lowbit(i)的意思是将i转化成二进制数之后,仅仅保留最低位的1及其后面的0,截断前面的内容,然后再转成10…

题目大意: 看一下样例就明白了 基本思路: 题目中明确提到k为一个周期,稍作思考,把k项看作一项,然后发现这是个等比数列,q=(b/a)^k, 然后重点就是怎样处理等比数列求和表达式中的除法,这个时候就要用到逆元,因为1e9+9是素数, 所以直接用费马小定理求逆元就好了,说到这个,可以学一下卢卡斯定理,这个比较有用处,然后需要注意 两点: 1)快速幂a的每次乘方里面都要%mod,这个到底是为什么我也不知道,难道不是只在外面取模一次就好了吗 2)最后判断条件是t2是否等于0,而不是a是否等于b,难…

1.HDU 5976 Detachment 2.题意:给一个正整数x,把x拆分成多个正整数的和,这些数不能有重复,要使这些数的积尽可能的大,输出积. 3.总结:首先我们要把数拆得尽可能小,这样积才会更大(当然不能拆1).所以容易想到是拆成2+3+...+n+s=x,先求出n即2+3+...+n<x<2+3+...+n+(n+1),然后将某个数向右平移s个单位变为n+1即可.注意:(1)预处理出前缀和,前缀积.(2)将某个数移到n+1,要除这个数再乘n+1,这里要用逆元,也要预处理出来. #in…

Codeforces Round #258 (Div. 2) Devu and Flowers E. Devu and Flowers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Devu wants to decorate his garden with flowers. He has purchased n boxes…

C16H:Magical Balls 总时间限制:  1000ms 内存限制:  262144kB 描述 Wenwen has a magical ball. When put on an infinite plane, it will keep duplicating itself forever. Initially, Wenwen puts the ball on the location (x0, y0) of the plane. Then the ball starts to dup…

3155: Preprefix sum Time Limit: 1 Sec  Memory Limit: 512 MBSubmit: 1183  Solved: 546[Submit][Status][Discuss] Description   Input 第一行给出两个整数N,M.分别表示序列长度和操作个数 接下来一行有N个数,即给定的序列a1,a2,....an 接下来M行,每行对应一个操作,格式见题目描述 Output 对于每个询问操作,输出一行,表示所询问的SSi的值. Sample…

ACdream 1214---矩阵连乘 Problem Description You might have noticed that there is the new fashion among rich people to have their yards tiled with black and white tiles, forming a pattern. The company Broken Tiles is well known as the best tiling company…

题意:求A^B的所有因子之和 很容易知道,先把分解得到,那么得到,那么 的所有因子和的表达式如下 第一种做法是分治求等比数列的和  用递归二分求等比数列1+pi+pi^2+pi^3+...+pi^n: (1)若n为奇数,一共有偶数项,则:      1 + p + p^2 + p^3 +...+ p^n = (1+p^(n/2+1)) + p * (1+p^(n/2+1)) +...+ p^(n/2) * (1+p^(n/2+1))      = (1 + p + p^2 +...+ p^(n/…

Read Phone Number Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu Submit Status Practice ACdream 1188 Description Do you know how to read the phone numbers in English? Now let me tell you. For example, In China, the phone n…

Alice and Bob Time Limit:3000MS     Memory Limit:128000KB     64bit IO Format:%lld & %llu Submit Status Practice ACdream 1112 Description Here  is Alice and Bob again ! Alice and Bob are playing a game. There are several numbers. First, Alice choose…

G - Happy 2004 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1452 Description Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to dete…

E - The Boss on Mars Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4059 Description On Mars, there is a huge company called ACM (A huge Company on Mars), and it’s owned by a younger boss. Due…

题目链接:http://acdream.info/problem?pid=1128 Problem Description wuyiqi陷入了一个迷宫中,这个迷宫是由N*M个格子组成的矩阵.每个格子上堆放了一定数量的箱子.(i,j)表示第i行,第j列的格子.wuyiqi可以将一个格子上的箱子移动到相邻的格子上,或者在这个格子上销毁.也就是在(i,j)的箱子可以移动到(i-1,j).(i+1,j).(i,j-1)和(i,j+1),但是不能移动到矩阵范围外.将(i,j)的格子上的一个箱子移动到相邻格…

hannnnah_j’s Biological Test Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 802    Accepted Submission(s): 269 Problem Description hannnnah_j is a teacher in WL High school who teaches biolog…

原题链接:http://acdream.info/problem?pid=1738 树套树裸题,如下: #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #define lc root<<1 #define rc root<<1|1 using std::sort; using std::l…

http://acdream.info/problem?pid=1726 官方题解:http://acdream.info/topic?tid=4246 求n个数里面能不能选一些数出来让它们的和等于k. 因为k很大,不能用背包,但是n很小,最大为40,所以拆成了2部分,之后最大为2^20次方<1050000;每次枚举前一半的和,然后用数组存储,然后得到一个总和减去后一半的差用二分查找. #include<cstdio> #include<cstring> #include&l…