D - Meaningless Sequence

最新推荐文章于 2024-07-27 23:39:44 发布
原创 最新推荐文章于 2024-07-27 23:39:44 发布 · 454 阅读 · 0 ·
CC 4.0 BY-SA版权
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。

这是一道关于数学和位操作的编程题目。题目要求求解一个特定的数列的和模10^9 + 7，数列定义涉及到位运算AND。你需要计算从0到n的累计和并取模。

https://codeforces.com/gym/102832/problem/D

Once there was a mathematician, who was obsessed with meaningless number sequences. Here is one of them.

 

an={1,c⋅max0≤i<nan&i,n=0otherwise,an={1,n=0c⋅max0≤i<nan&⁡i,otherwise,

where && denotes the bitwise AND operation.

As a mathematician, he could easily tell what anan was for any nn, but he wanted to test you. You are required to tell him

 

(∑i=0nai)mod(109+7)(∑i=0nai)mod(109+7)

to convince him that you have a deep understanding of this (although meaningless) sequence.

Input

The only line contains two integers nn and cc (0≤n<23000,0≤c≤109)0≤n<23000,0≤c≤109).

Specially, nn is given in binary presentation. cc is given in decimal presentation normally.

Output

最低0.47元/天 解锁文章
【2020ccpc长春站】D题 Meaningless Sequence（dp，数学） 详解
Qiuker_jl的博客
11-12 753
题目链接：Meaningless Sequence
2020 ccpc长春 D. Meaningless Sequence(按位启发式合并)
jziwjxjd的博客
05-24 491
LINK 刚好在gymgymgym碰到就顺手写了 话说打长春的时候是被队友秒掉的emm 自己写的时候还是有点细节的… 注意虽然这是一颗有根树,但边并不是完全由父亲指向儿子的!!! 需要建双向边,我被这个卡疯了… ∑i=1n∑j=i+1n[ai⊕aj=alcai,j]∗(i⊕j)\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n[a_i\oplus a_j=a_{lca_{i,j}}]*(i\oplus j)i=1∑n​j=i+1∑n​[ai​⊕aj​=alcai,j​​]∗(
2020CCPC长春D. Meaningless Sequence(思维+数位DP）
Buyi的博客
12-01 634
题目链接：http://codeforces.com/gym/102832/problem/D 解题思路： 可以发现一个数的大小与它的二进制表示中的1的个数有关， a=c^(二进制中1的个数） 那么题目就转化为求所有数中1的个数 使用的是数位dp的方法，枚举1的个数来分配。 对于没有上限要求的x长度串中分配y个1的方案数直接可以使用组合数C(y,x) #include<iostream> #include<cstdio> #include<string.h> #inclu
2020CCPC长春 7-4 D - Meaningless Sequence（数位DP）
tomjobs的博客
11-12 501
思路： 可以发现，ana_nan​的二进制有xxx个1，则值为cxc^xcx，所以我们要求aia_iai​二进制中1的个数。 定义dp[i][j]dp[i][j]dp[i][j]为考虑i~n位，有j个1的情况有多少种，然后裸的数位DP跑就好了。 好像直接固定前缀就可以n^2写了，数位dp并没有必要。。。 #include <cstdio> #include <algorithm> #include <cstring> using namespace std; con.
D - Meaningless Sequence 2020CCPC长春
binarycopycode
11-08 348
a_t=c^(pop_count*(t)) 傻逼数位DP，这题我竟然写了半小时？身败名裂 #include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxl=3010; const int mod=1e9+7; int n,c;ll ans; ll dp[maxl][2]; char s[maxl]; inline void prework() { scanf("%s",s+1); n
2020 CCPC 长春 D. Meaningless Sequence(思维+二进制)
chr1stopher
11-13 358
Description Example input 1000 3 output 67 Solution 容易发现ai=cka_i = c^kai​=ck(kkk为iii的二进制下1的个数)，我们枚举1的位置来判断所有可能值即可统计答案 Code #include <bits/stdc++.h> #define pb push_back typedef long long ll; using namespace std; const int mod = 1e9 + 7; const int ma
《Utilizing Ensemble Learning for Detecting Multi-Modal Fake News》
buyaotutou的博客
07-27 846
文章链接利用集成学习检测多模态假新闻The spread of fake news has become a critical problem in recent years due extensive use of social media platforms. False stories can go viral quickly, reaching millions of people before they can be mocked, i.e., a false story claiming tha
CMU 11-785 L16 Connectionist Temporal Classification
crazy_scott的博客
08-06 312
Sequence to sequence Sequence goes in, sequence comes out No notion of “time synchrony” between input and output May even nots maintain order of symbols (from one language to another) With order synchrony The input and output sequences happen in the
《Multi-Modal Features Representation-Based Convolutional Neural Network Model for Malicious Website》
buyaotutou的博客
07-27 635
文章链接基于多模态特征表示的卷积神经网络恶意网站检测模型Web applications have proliferated across various business sectors, serving as essential tools for billions of users in their daily lives activities. However, many of these applications are malicious which is a major threat to I
2020年ccpc长春站D-Meaningless Sequence
林苏泽
11-09 710
首先引用一下大佬的博客 推一下为什么是这样的： 首先：当二进制是满1的话就是∑i=0xai=(1+c)n\sum_{i=0}^{x}a_{i} = (1+c)^{n}∑i=0x​ai​=(1+c)n其中nnn为xxx的二进制位数，这个其实很容易就能看出来吧按我的思路说一下：相当于一共nnn位在nnn位取小于等于nnn个1其实就是组合数Cn0∗c0+Cn1∗c1+...+Cnn∗cn=(1+c)nC_{n}^{0}*c^{0}+C_{n}^{1}*c^{1}+...+C_{n}^{n}*c^{n} = (1.
视频-分布式链路追踪skywalking.txt
10-24
视频-分布式链路追踪skywalking.txt
2020ccpc长春D Meaningless Sequence（推规律 nlogn 解法）
羊驼的博客
11-08 1770
赛场交了个 O(nlogn)O(nlogn)O(nlogn) 的代码，居然过了 第一次现场赛碾压标程（雾） 补一下想法： 一开始发现了一个规律，对于一个二进制位全为 111 的数 xxx 有 ∑i=0xai=(1+c)n\sum_{i=0}^xa_i=(1+c)^n∑i=0x​ai​=(1+c)n 其中 nnn 为 xxx 的二进制位数，而当 xxx 的二进制第一位是 111 其余都是 000 的数时有 ∑i=0xai=(1+c)n+c\sum_{i=0}^xa_i=(1+c)^n+c∑i=0x​a.
C. Infinite Fence
Luoriliming的博客
11-28 1024
链接：https://codeforces.com/contest/1260/problem/C You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your pun...
A. Plate Game
Luoriliming的博客
01-18 725
链接：https://codeforces.com/problemset/problem/197/A You've got a rectangular table with lengthaand widthband the infinite number of plates of radiusr. Two players play the following game: they ta...
C 装备合成
Luoriliming的博客
03-14 644
链接：https://ac.nowcoder.com/acm/contest/4743/C 牛牛有{x}x件材料{a}a和{y}y件材料{b}b，用{2}2件材料{a}a和{3}3件材料{b}b可以合成一件装备，用{4}4件材料{a}a和{1}1件材料{b}b也可以合成一件装备。牛牛想要最大化合成的装备的数量，于是牛牛找来了你帮忙。 输入描述: 输入包含{t}t组数据 第一行一个整数{t}...
D. Same GCDs
Luoriliming的博客
01-30 641
链接：https://codeforces.com/contest/1295/problem/D You are given two integersaaandmm. Calculate the number of integersxxsuch that0≤x<m0≤x<mandgcd(a,m)=gcd(a+x,m)gcd(a,m)=gcd(a+x,m). Note...
D. Count the Arrays
Luoriliming的博客
03-10 616
链接：https://codeforces.ml/contest/1312/problem/D Your task is to calculate the number of arrays such that: each array containsnnelements; each element is an integer from11tomm; for each array,...
C. Banh-mi
Luoriliming的博客
01-22 587
链接：https://codeforces.com/group/nhsfFrN6WT/contest/266629/problem/C JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as ...
D. Aroma's Search
Luoriliming的博客
01-20 573
链接：https://codeforces.com/contest/1293/problem/D THE SxPLAY & KIVΛ - 漂流 KIVΛ & Nikki Simmons - Perspectives With a new body, our idol Aroma White (or should we call her Kaori Minamiya?) beg...
这里呢？ LOCAL STATUS https_handle(int sock) {HTTP_ERROR("进入https_handle"); CONTEXT *context = NULL; #ifdef OPTEE_SUPPORT TEEC_Result ret = TEEC_SUCCESS; unsigned char priv_key_buf[TEE_STORAGE_MAX_LEN] = {0}; UINT32 priv_key_len = 0; unsigned char priv_rsa_key_buf[TEE_STORAGE_MAX_LEN] = {0}; UINT32 priv_rsa_key_len = 0; struct tee_client_ctx ctx; TEEC_UUID uuid = TA_SECURE_STORAGE_UUID; #endif if ((context = socket_handle(sock)) == NULL) return ERROR; context->is_https = TRUE; #ifdef OPTEE_SUPPORT memset(&ctx, 0x0, sizeof(ctx)); memset(&priv_key_buf[0], 0x0, sizeof(priv_key_buf)); memset(&priv_rsa_key_buf[0], 0x0, sizeof(priv_rsa_key_buf)); ret = lib_tee_cli_prepare_tee_session(&ctx, uuid, LIB_TEE_CLI_TA_SECURE_STORAGE); if(TEEC_SUCCESS != ret) { HTTPD_ERROR("open tee ctx fail!\n"); } ret |= lib_tee_cli_mem_manage(&ctx, LIB_TEE_CLI_MEM_HTTPD_ECC_PRIV_KEY, (char *)&priv_key_buf[0], priv_key_len, TEE_MEM_MANAGE_MODE_READ); ret |= lib_tee_cli_mem_manage(&ctx, LIB_TEE_CLI_MEM_HTTPD_RSA_PRIV_KEY, (char *)&priv_rsa_key_buf[0], priv_rsa_key_len, TEE_MEM_MANAGE_MODE_READ); if(TEEC_SUCCESS != ret) { HTTPD_ERROR("tee READ fail!\n"); ret = lib_tee_cli_terminate_tee_session(&ctx, LIB_TEE_CLI_TA_SECURE_STORAGE); context_free(context); HTTPD_ERROR("lib_tee_cli_terminate_tee_session %d\n",ret); return ERROR; } // two void functions are called in lib_tee_cli_terminate_tee_session. ret is meaningless. ret = lib_tee_cli_terminate_tee_session(&ctx, LIB_TEE_CLI_TA_SECURE_STORAGE); HTTPD_WARNING("lib_tee_cli_terminate_tee_session %d\n",ret); priv_key_len = strlen((char *)priv_key_buf); priv_rsa_key_len = strlen((char *)priv_rsa_key_buf); HTTPD_WARNING("read success! priv_key_buf len %d\n%s\n, priv_rsa_key_buf len %d\n%s", priv_key_len, priv_key_buf, priv_rsa_key_len, priv_rsa_key_buf); priv_key_buf[priv_key_len] = '\0'; priv_rsa_key_buf[priv_rsa_key_len] = '\0'; if(0 == access(HTTP_CERT_TMP_PATH, F_OK)) { context->ssl_ctx = tpssl_svr_init(HTTP_CERT_TMP_PATH, (char *)&priv_key_buf[0], 0, SERVER_WRITE_CERT_USING_SUBJECT_KEY, TRUE, (char *)&priv_rsa_key_buf[0]); } else { HTTPD_ERROR("No cert!\n",ret); return ERROR; } #else //context->ssl_ctx = tpssl_svr_init(HTTP_CERT_PATH, HTTP_KEY_PATH, 0); if(0 == access(HTTP_CERT_TMP_PATH, F_OK) && 0 == access(HTTP_KEY_TMP_PATH, F_OK)) { context->ssl_ctx = tpssl_svr_init(HTTP_CERT_TMP_PATH, HTTP_KEY_TMP_PATH, 0, SERVER_WRITE_CERT_USING_SUBJECT_KEYFILE, FALSE, NULL); } else { context->ssl_ctx = tpssl_svr_init(HTTP_CERT_PATH, HTTP_KEY_PATH, 0, SERVER_WRITE_CERT_USING_SUBJECT_KEYFILE, FALSE, NULL); } #endif tpssl_svr_set_bio(context->ssl_ctx, context->sock); //握手失败，不应该再对该session进行后续接收或发送处理，否则会出现接收或发送失败，产生不必要的打印 //web由于证书自签，第一次握手不会成功 if (tpssl_svr_handshake(context->ssl_ctx) == -1) { context_free(context); return ERROR; } match_group(context, sock); return OK; }
最新发布
12-27
sequenceDiagram 浏览器->>IPC摄像头: POST /set_static_ip IPC摄像头->>TEE: 打开安全存储会话(prepare) TEE-->>IPC摄像头: 成功响应 IPC摄像头->>网络服务: 重启网络接口 网络服务->>IPC摄像头: 关闭网络接口...
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
表情包 代码片
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付：10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明：

1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

余额充值