CodeForces 221(div 2)
A
无trick水题。。。
/*
* Author: Plumrain
* Created Time: 2013-12-24 22:26
* File Name: B.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <fstream>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cctype>
#include <ctime>
#include <utility> using namespace std; #define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define INF 999999999999999999
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false) const double eps = 1e-;
const double PI = atan(1.0)*;
const int inf = / ; typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef long long int64; inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;} int ru[], chu[]; int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
// std::ios::sync_with_stdio(false);
int n, m;
while (scanf ("%d%d", &n , &m) != EOF){
int t1, t2, w;
clr0 (chu); clr0 (ru);
for (int i = ; i < m; ++ i){
scanf ("%d%d%d", &t1, &t2, &w);
chu[--t1] += w;
ru[--t2] += w;
}
int ans = ;
for (int i = ; i < n; ++ i)
ans += abs(chu[i] - ru[i]);
printf ("%d\n", ans / );
}
return ;
}
B
YY题。。。结论直接看代码就好了。。。比赛的时候YY出了结论算样例算错了,然后就去想别的方法了。。。。。。
/*
* Author: Plumrain
* Created Time: 2013-12-24 22:26
* File Name: B.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <fstream>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cctype>
#include <ctime>
#include <utility> using namespace std; #define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define INF 999999999999999999
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false) const double eps = 1e-;
const double PI = atan(1.0)*;
const int inf = / ; typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef long long int64; inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;} int ru[], chu[]; int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
// std::ios::sync_with_stdio(false);
int n, m;
while (scanf ("%d%d", &n , &m) != EOF){
int t1, t2, w;
clr0 (chu); clr0 (ru);
for (int i = ; i < m; ++ i){
scanf ("%d%d%d", &t1, &t2, &w);
chu[--t1] += w;
ru[--t2] += w;
}
int ans = ;
for (int i = ; i < n; ++ i)
ans += abs(chu[i] - ru[i]);
printf ("%d\n", ans / );
}
return ;
}
C
题意:给一个很大的数m,这个数的各个位上的数字中一定含有至少1个1,6,8,9。你可以重新组织所有数字的顺序,使得重新排列之后的数能被7整除。10^4 <= m <= 10^(10^6)。
注意,排列之后的数不能含有前导0。
解法:把1,6,8,9四个数字放在最后面,根据前面的数*10000除以7的余数,来决定1,6,8,9四个数字的排列顺序即可。至于有没有前导0,特殊处理一下即可。
tag:math, think
/*
* Author: Plumrain
* Created Time: 2013-12-25 14:09
* File Name: C.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <fstream>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cctype>
#include <ctime>
#include <utility> using namespace std; #define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define INF 999999999999999999
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false) const double eps = 1e-;
const double PI = atan(1.0)*;
const int inf = / ; typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef long long int64; inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;} bool del[];
string temp = "";
map<int, string> mp; void gao(string s)
{
stringstream stm(s);
int num; stm >> num;
int yu = num % ;
if (!mp.count(yu)) mp[yu] = s;
} int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
// std::ios::sync_with_stdio(false);
mp.clear();
string tt = "";
gao(tt);
while (next_permutation(tt.begin(), tt.end())) gao(tt); string s;
while (cin >> s){
clr0 (del);
string ss; ss.clear();
int n = sz(s), cnt = ;
for (int i = ; i < n; ++ i){
if (s[i] == ''){
++ cnt; continue;
}
bool ok = ;
for (int j = ; j < ; ++ j) if (s[i] == temp[j] && !del[j]){
del[j] = ; ok = ;
}
if (!ok) ss.pb (s[i]);
} int len = sz(ss);
if (!len){
ss = mp[];
for (int i = ; i < cnt; ++ i) ss.pb ('');
cout << ss << endl;
continue;
}
for (int i = ; i < cnt; ++ i) ss.pb ('');
len = sz(ss);
int flag = ;
for (int i = ; i < len; ++ i)
flag = (flag* + ss[i] - '') % ;
flag = flag * % ;
ss += mp[( - flag) % ];
cout << ss << endl;
}
return ;
}
D
题意:有一个0,1矩阵(最大5000*5000),你可以无限次数地交换任意两行的位置。求交换之后,单个只含有1的矩形的面积最大,并返回这个面积值。
解法:枚举矩形的左下角是从哪一列开始,统计每一行从这一列开始连续的1有多少个记录在num数组里,然后从大到小遍历num数组,并更新面积值即可。
tag:dp, think, good
/*
* Author: Plumrain
* Created Time: 2013-12-26 12:49
* File Name: D.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <fstream>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cctype>
#include <ctime>
#include <utility> using namespace std; #define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define INF 999999999999999999
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false) const double eps = 1e-;
const double PI = atan(1.0)*;
const int inf = / ; typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef long long int64; inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;} int n, m;
int p[][], num[];
char v[][]; int main()
{
// freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
// std::ios::sync_with_stdio(false);
while (scanf ("%d%d", &n, &m) != EOF){
string s;
for (int i = ; i < n; ++ i)
scanf ("%s", v[i]);
for (int i = ; i < n; ++ i){
int tmp = m;
for (int j = m-; j >= ; -- j){
if (v[i][j] == '')
tmp = j, p[i][j] = j;
else
p[i][j] = tmp;
}
}
int ans = ;
for (int i = ; i < m; ++ i){
clr0 (num);
for (int j = ; j < n; ++ j) num[p[j][i] - i] ++;
int pos = ;
while (pos && num[pos] == ) -- pos;
int cnt = ;
while (pos){
if (num[pos]) cnt += num[pos];
ans = max(cnt*pos, ans);
-- pos;
}
}
printf ("%d\n", ans);
}
return ;
}
CodeForces 221(div 2)的更多相关文章
- Codeforces #344 Div.2
Codeforces #344 Div.2 Interview 题目描述:求两个序列的子序列或操作的和的最大值 solution 签到题 时间复杂度:\(O(n^2)\) Print Check 题目 ...
- Codeforces #345 Div.1
Codeforces #345 Div.1 打CF有助于提高做题的正确率. Watchmen 题目描述:求欧拉距离等于曼哈顿距离的点对个数. solution 签到题,其实就是求有多少对点在同一行或同 ...
- Codeforces Beta Round #27 (Codeforces format, Div. 2)
Codeforces Beta Round #27 (Codeforces format, Div. 2) http://codeforces.com/contest/27 A #include< ...
- Codeforces#441 Div.2 四小题
Codeforces#441 Div.2 四小题 链接 A. Trip For Meal 小熊维尼喜欢吃蜂蜜.他每天要在朋友家享用N次蜂蜜 , 朋友A到B家的距离是 a ,A到C家的距离是b ,B到C ...
- codeforces #592(Div.2)
codeforces #592(Div.2) A Pens and Pencils Tomorrow is a difficult day for Polycarp: he has to attend ...
- codeforces #578(Div.2)
codeforces #578(Div.2) A. Hotelier Amugae has a hotel consisting of 1010 rooms. The rooms are number ...
- codeforces #577(Div.2)
codeforces #577(Div.2) A Important Exam A class of students wrote a multiple-choice test. There are ...
- codeforces #332 div 2 D. Spongebob and Squares
http://codeforces.com/contest/599/problem/D 题意:给出总的方格数x,问有多少种不同尺寸的矩形满足题意,输出方案数和长宽(3,5和5,3算两种) 思路:比赛的 ...
- Codeforces Round #221 (Div. 1) B. Maximum Submatrix 2 dp排序
B. Maximum Submatrix 2 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset ...
随机推荐
- asp.net 实现对xml文件的 读取,添加,删除,修改
用于修改站内xml文件 已知有一个XML文件(bookstore.xml)如下:<?xml version="1.0" encoding="gb2312" ...
- 11g 创建asm磁盘组
[root@Oracle11g ~]# fdisk -l Disk /dev/sda: 21.4 GB, 21474836480 bytes255 heads, 63 sectors/track, 2 ...
- kvc简单实现
除了一般的赋值和取值的方法,我们还可以用Key-Value-Coding(KVC)键值编码来访问你要存取的类的属性 kvc: kvc key value coding 键值对编码 可以通过 ...
- 读懂IL代码(二)
上一篇提到了最基本的IL代码,应该是比较通俗易懂的,所以有了上一篇的基础之后,这篇便要深入一点点的来讲述了. 首先我必须再来说一些重要的概念: Evaluation Stack(评估栈):这是由.NE ...
- jquery中eq和get的区别与使用方法
$("p").eq(0).css("color") //因为eq(num)返回的是个jq对象,所以可以用jq的方法css使用get来获得第一个p标签的color ...
- Tomcat环境变量的配置
Tomcat web服务器 支持全部JSP以及Servlet规范 主要作用 是提供了一个可以让Jsp和Servlet运行的平台 tomcat环境变量设置 CATALINA_HOME : D:\bran ...
- C#.NET快速开发框架-企业版V4.0截图打包下载
C/S系统开发框架-企业版 V4.0 (Enterprise Edition) http://www.csframework.com/cs-framework-4.0.htm 其它图片打包下载: ht ...
- jquery如何判断div是否隐藏--useful
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...
- linux中硬盘及网卡的表示方法
Linux中的所有设备均表示为/dev下的一个文件,各种IDE设备分配一个由hd前缀组成的文件:而对于各种SCSI设备,则分配了一个由sd前缀组成的文件,例如: IDE0接口上的主盘成为/dev/hd ...
- c++:参数型别的推导
STL源码剖析--侯捷 总结 尽管现在的很多语言支持参数类型的判别,但是c/c++并不支持这一特性. 但是我们可以通过一些技巧使得c++具有自动判别参数类型的特性. 模板 我们都知道在模板类和模板函数 ...