PAT/简单模拟习题集（二）
B1018. 锤子剪刀布 (20)
Discription：
大家应该都会玩“锤子剪刀布”的游戏：两人同时给出手势，胜负规则如图所示：
现给出两人的交锋记录，请统计双方的胜、平、负次数，并且给出双方分别出什么手势的胜算最大。
Input：
输入第1行给出正整数N（<=105），即双方交锋的次数。随后N行，每行给出一次交锋的信息，即甲、乙双方同时给出的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”，第1个字母代表甲方，第2个代表乙方，中间有1个空格。
Output：
输出第1、2行分别给出甲、乙的胜、平、负次数，数字间以1个空格分隔。第3行给出两个字母，分别代表甲、乙获胜次数最多的手势，中间有1个空格。如果解不唯一，则输出按字母序最小的解。
Sample Input：
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J
Sample Output：
5 3 2
2 3 5
B B
#include <cstdio> #define MaxSize 3 int ListA[MaxSize], ListB[MaxSize], Time[MaxSize]; int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int N;
char c1, c2;
scanf("%d", &N);
for(int i=; i<N; i++) {
getchar();
scanf("%c %c", &c1, &c2);
if(c1 == 'B') {
if(c2 == 'B') {
Time[]++;
} else if(c2 == 'C') {
Time[]++;
ListA[]++;
} else {
Time[]++;
ListB[]++;
}
} else if(c1 == 'C') {
if(c2 == 'B') {
Time[]++;
ListB[]++;
} else if(c2 == 'C') {
Time[]++;
} else {
Time[]++;
ListA[]++;
}
} else {
if(c2 == 'B') {
Time[]++;
ListA[]++;
} else if(c2 == 'C') {
Time[]++;
ListB[]++;
} else {
Time[]++;
}
}
} printf("%d %d %d\n%d %d %d\n", Time[], Time[], Time[], Time[], Time[], Time[]);
if(ListA[] >= ListA[]) {
if(ListA[] >= ListA[])
printf("B ");
else
printf("J ");
} else {
if(ListA[] >= ListA[])
printf("C ");
else
printf("J ");
}
if(ListB[] >= ListB[]) {
if(ListB[] >= ListB[])
printf("B\n");
else
printf("J\n");
} else {
if(ListB[] >= ListB[])
printf("C\n");
else
printf("J\n");
} return ;
}
#include <cstdio> int change(char c)
{
if(c == 'B')
return ;
else if(c == 'C')
return ;
else
return ;
} int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); char mp[] = {'B', 'C', 'J'};
int n;
scanf("%d", &n);
int times_A[] = {}, times_B[] = {};
int hand_A[] = {}, hand_B[] = {};
char c1, c2;
int k1, k2;
for(int i=; i<n; i++) {
getchar();
scanf("%c %c", &c1, &c2);
k1 = change(c1);
k2 = change(c2);
if((k1+)% == k2) {
times_A[]++;
times_B[]++;
hand_A[k1]++;
} else if(k1 == k2) {
times_A[]++;
times_B[]++;
} else {
times_A[]++;
times_B[]++;
hand_B[k2]++;
}
} printf("%d %d %d\n", times_A[], times_A[], times_A[]);
printf("%d %d %d\n", times_B[], times_B[], times_B[]);
int id1 = , id2 = ;
for(int i=; i<; i++) {
if(hand_A[i] > hand_A[id1])
id1 = i;
if(hand_B[i] > hand_B[id2])
id2 = i;
}
printf("%c %c\n", mp[id1], mp[id2]); return ;
}
A1042. Shuffling Machine (20)
Description：
Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid "inside jobs" where employees collaborate with gamblers by performing inadequate shuffles, many casinos employ automatic shuffling machines. Your task is to simulate a shuffling machine.
The machine shuffles a deck of 54 cards according to a given random order and repeats for a given number of times. It is assumed that the initial status of a card deck is in the following order:
S1, S2, ..., S13, H1, H2, ..., H13, C1, C2, ..., C13, D1, D2, ..., D13, J1, J2
where "S" stands for "Spade", "H" for "Heart", "C" for "Club", "D" for "Diamond", and "J" for "Joker". A given order is a permutation of distinct integers in [1, 54]. If the number at the i-th position is j, it means to move the card from position i to position j. For example, suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we are to repeat the shuffling again, the result will be: C1, H5, S3, J2, D13.
Input：
Each input file contains one test case. For each case, the first line contains a positive integer K (<= 20) which is the number of repeat times. Then the next line contains the given order. All the numbers in a line are separated by a space.
Output：
For each test case, print the shuffling results in one line. All the cards are separated by a space, and there must be no extra space at the end of the line.
Sample Input：
2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47
Sample Output：
S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5
#include <cstdio> #define MaxSize 55 int List1[MaxSize], List2[MaxSize], List3[MaxSize]; int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int N, temp;
scanf("%d", &N);
for(int i=; i<MaxSize; i++) {
List1[i] = i;
scanf("%d", &List2[i]);
}
for(int i=; i<N; i++) {
for(int j=; j<MaxSize; j++) {
List3[List2[j]] = List1[j];
}
for(int k=; k<MaxSize; k++) {
List1[k] = List3[k];
}
} for(int i=; i<MaxSize-; i++) {
if(List1[i] > ) {
printf("J%d ", List1[i]-);
} else if(List1[i] > ) {
printf("D%d ", List1[i]-);
} else if(List1[i] > ) {
printf("C%d ", List1[i]-);
} else if(List1[i] > ) {
printf("H%d ", List1[i]-);
} else {
printf("S%d ", List1[i]);
}
}
if(List1[MaxSize-] > ) {
printf("J%d\n", List1[MaxSize-]-);
} else if(List1[MaxSize-] > ) {
printf("D%d\n", List1[MaxSize-]-);
} else if(List1[MaxSize-] > ) {
printf("C%d\n", List1[MaxSize-]-);
} else if(List1[MaxSize-] > ) {
printf("H%d\n", List1[MaxSize-]-);
} else {
printf("S%d\n", List1[MaxSize-]);
} return ;
}
#include <cstdio> const int N = ;
char mp[] = {'S', 'H', 'C', 'D', 'J'};
int start[N], end[N], next[N]; int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int K;
scanf("%d", &K);
for(int i=; i<=N; i++)
start[i] = i;
for(int i=; i<=N; i++)
scanf("%d", &next[i]); for(int step = ; step < K; step++) {
for(int i=; i<=N; i++)
end[next[i]] = start[i];
for(int i=; i<=N; i++)
start[i] = end[i];
} for(int i=; i<=N; i++) {
if(i != )
printf(" ");
start[i]--;
printf("%c%d", mp[start[i]/], start[i]%+);
} return ;
}
A1046.Shortest Distance (20)
Description：
The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.
Input：
Each input file contains one test case. For each case, the first line contains an integer N (in [3, 105]), followed by N integer distances D1 D2 ... DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (<=104), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 107.
Output：
For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.
Sample Input：
5 1 2 4 14 9
3
1 3
2 5
4 1
Sample Output：
3
10
7
#include <cstdio>
#include <algorithm>
using namespace std; const int MAXN = ;
int dis[MAXN], A[MAXN]; int main()
{
int sum = , query, n, left, right;
scanf("%d", &n);
for(int i=; i<=n; i++) {
scanf("%d", &A[i]);
sum += A[i];
dis[i] = sum;
}
scanf("%d", &query);
for(int i=; i<query; i++) {
scanf("%d%d", &left, &right);
if(left > right)
swap(left, right);
int temp = dis[right-]-dis[left-];
printf("%d\n", min(temp, sum-temp));
} return ;
}
A1065. A+B and C (64bit) (20)
Description：
Given three integers A, B and C in [-263, 263], you are supposed to tell whether A+B > C.
Input：
The first line of the input gives the positive number of test cases, T (<=10). Then T test cases follow, each consists of a single line containing three integers A, B and C, separated by single spaces.
Ouput：
For each test case, output in one line "Case #X: true" if A+B>C, or "Case #X: false" otherwise, where X is the case number (starting from 1).
Sample Input：
3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0
Output：
Case #1: false
Case #2: true
Case #3: false
#include <cstdio> int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int T, tcase = ;
scanf("%d", &T);
while(T--) {
long long a, b, c;
scanf("%lld%lld%lld", &a, &b, &c);
long long res = a+b;
bool flag;
if(a> && b> && res<)
flag = true;
else if(a< && b< && res>=)
flag = false;
else if(res > c)
flag = true;
else
flag = false;
if(flag == true)
printf("Case #%d: true\n", tcase++);
else
printf("Case #%d: false\n", tcase++);
} return ;
}
B1010. 一元多项式求导 (25)
Description：
设计函数求一元多项式的导数。（注：xn（n为整数）的一阶导数为n*xn-1。）
Input：
以指数递降方式输入多项式非零项系数和指数（绝对值均为不超过1000的整数）。数字间以空格分隔。
Output：
以与输入相同的格式输出导数多项式非零项的系数和指数。数字间以空格分隔，但结尾不能有多余空格。注意“零多项式”的指数和系数都是0，但是表示为“0 0”。
Sample Input：
3 4 -5 2 6 1 -2 0
Sample Output：
12 3 -10 1 6 0
#include <cstdio> int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int a[] = {};
int k, e, counter = ;
while(scanf("%d%d", &k, &e) != EOF) {
a[e] = k;
}
a[] = ; for(int i=; i<=; ++i) {
a[i-] = a[i]*i;
a[i] = ;
if(a[i-] != )
counter++;
}
if(counter == )
printf("0 0\n");
else {
for(int i=; i>=; --i) {
if(a[i] != ) {
printf("%d %d", a[i], i);
counter--;
if(counter != )
printf(" ");
}
}
} return ;
}
A1002. A+B for Polynomials (25)
Description：
This time, you are supposed to find A+B where A and B are two polynomials.
Input：
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.
Output：
For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
Sample Input：
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output：
3 2 1.5 1 2.9 0 3.2
#include <cstdio> #define MaxSize 1010
double List[MaxSize]; int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int K, expon, counter = ;
double coef;
scanf("%d", &K);
for(int i=; i<K; ++i) {
scanf("%d %lf", &expon, &coef);
List[expon] += coef;
}
scanf("%d", &K);
for(int i=; i<K; ++i) {
scanf("%d %lf", &expon, &coef);
List[expon] += coef;
} for(int i=; i<MaxSize; ++i) {
if(List[i] != )
++counter;
}
printf("%d", counter);
for(int i=MaxSize-; i>=; --i) {
if(List[i] != ) {
printf(" %d %.1f", i, List[i]);
}
} return ;
}
A1009. Product of Polynomials (25)
Description：
This time, you are supposed to find A*B where A and B are two polynomials.
Input：
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.
Output：
For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
Sample Input：
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output：
3 3 3.6 2 6.0 1 1.6
#include <cstdio> #define MaxSize 2010
double List1[MaxSize], List2[MaxSize]; int main()
{
//freopen("E:\\Temp\\input.txt", "r", stdin); int K, expon, counter = ;
double coef;
scanf("%d", &K);
for(int i=; i<K; ++i) {
scanf("%d %lf", &expon, &coef);
List1[expon] += coef;
}
scanf("%d", &K);
for(int i=; i<K; ++i) {
scanf("%d %lf", &expon, &coef);
for(int j=; j<MaxSize; j++)
List2[expon+j] += List1[j]*coef;
} for(int i=; i<MaxSize; ++i) {
if(List2[i] != )
++counter;
}
printf("%d", counter);
for(int i=MaxSize-; i>=; --i) {
if(List2[i] != )
printf(" %d %.1f", i, List2[i]);
} return ;
}
#include <cstdio>
struct Poly {
int exp;
double cof;
}poly[];
double ans[];
int main()
{
int n, m, number = ;
scanf("%d", &n);
for(int i=; i<n; ++i)
scanf("%d %lf", &poly[i].exp, &poly[i].cof);
scanf("%d", &m);
for(int i=; i<m; ++i) {
int exp;
double cof;
scanf("%d %lf", &exp, &cof);
for(int j=; j<n; j++)
ans[exp+poly[j].exp] += (cof*poly[j].cof);
}
for(int i=; i<=; ++i) {
if(ans[i] != )
++number;
}
printf("%d", number);
for(int i=; i>=; --i) {
if(ans[i] != )
printf(" %d %.1f", i, ans[i]);
}
return ;
}
PAT/简单模拟习题集（二）的更多相关文章
- PAT/简单模拟习题集（一）
B1001.害死人不偿命的(3n+1)猜想 (15) Description: 卡拉兹(Callatz)猜想: 对任何一个自然数n,如果它是偶数,那么把它砍掉一半:如果它是奇数,那么把(3n+1)砍掉 ...
- java web学习总结(二十二) -------------------简单模拟SpringMVC
在Spring MVC中,将一个普通的java类标注上Controller注解之后,再将类中的方法使用RequestMapping注解标注,那么这个普通的java类就够处理Web请求,示例代码如下: ...
- JavaWeb学习总结(四十九)——简单模拟Sping MVC
在Spring MVC中,将一个普通的java类标注上Controller注解之后,再将类中的方法使用RequestMapping注解标注,那么这个普通的java类就够处理Web请求,示例代码如下: ...
- (hdu step 8.1.6)士兵队列训练问题(数据结构，简单模拟——第一次每2个去掉1个,第二次每3个去掉1个.知道队伍中的人数<=3，输出剩下的人 )
题目: 士兵队列训练问题 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...
- spring之mvc原理分析及简单模拟实现
在之前的一篇博客中已经简单的实现了spring的IOC和DI功能,本文将在之前的基础上实现mvc功能. 一 什么是MVC MVC简单的说就是一种软件实现的设计模式,将整个系统进行分层,M(model ...
- C++笔记（7）——一些模拟题：简单模拟、查找元素、图形输出、日期处理、进制转换、字符串处理
以下内容基本来自<算法笔记>,作者为胡凡,建议直接买书看,我这里只是摘抄部分当笔记,不完整的. 简单模拟 就是一类"题目怎么说你就怎么做"的题目.这类题目不涉及算法,只 ...
- sort回调的简单模拟
本来是准备讲CPP中的std::sort,但因为最近Java用得多,不知怎么的便习惯性走Java角度看问题了,所以这篇文章看起来估计会有点奇怪... 一.简单模拟sort回调 std::sort函数本 ...
- WPF简单模拟QQ登录背景动画
介绍 之所以说是简单模拟,是因为我不知道QQ登录背景动画是怎么实现的.这里是通过一些办法把它简化了,做成了类似的效果 效果图 大体思路 首先把背景看成是一个4行8列的点的阵距,X轴Y轴都是距离70.把 ...
- Linux 内核 链表 的简单模拟（2）
接上一篇Linux 内核 链表 的简单模拟(1) 第五章:Linux内核链表的遍历 /** * list_for_each - iterate over a list * @pos: the & ...
随机推荐
- JS execCommand 方法
document.execCommand()方法处理Html数据时常用语法格式如下: 复制内容到剪贴板 代码: document.execCommand(sCommand[,交互方式, 动态参数]) ...
- android中ColorStateList及StateListDrawable设置Selector
写过android的代码相信大家对Selector并不陌生吧,下面来看看这段xml文件是如何定义的 <?xml version="1.0" encoding="ut ...
- Robot Framework入门学习2 创建第一个测试用例
本文章部分内容引自以下网址,感谢作者的辛苦分享 http://www.cnblogs.com/fnng/p/3871712.html http://blog.csdn.net/tulituqi/art ...
- MySQL之运算符与函数、自定义函数
一自定义函数简介 (1)自定义函数定义 用户自定义函数(user-defined function,UDF)是一种对MySQL数据库扩展的途径,其用法与内置函数相同. (2)自定义函数的两个必要条件 ...
- postgreSQL 统计语句
pg_stat_statements 是 postgresql 的一个扩展,用来统计查询语句,类似于 mysql 的 慢查询. 安装二进制文件 有些发行版可能没有附带这个扩展,则需要用户自己安装, 本 ...
- 交换机的link-dependency链路依赖功能
在生产环境中,如果各主机连接到交换机的1-16端口,交换机的18端口上联到外部网络,如果此时交换机的18端口断掉,在主机端网卡仍然有连接,对于一些网络的配置可能就感知不到外部连接的消失,会有问题(如E ...
- 深入研究C语言 第一篇
一． 研究过程 1.第一章:创建编译环境: 我们首先下载TC2.0,找到其中与编译连接相关的程序和文件: (1) 编译器:TCC.exe (2) 连接器:tllike.exe (3) 相关文件:c0s ...
- idea 从github下载项目提示 file name too long 的解决方案
1.找到git shell命令行 2运行如下命令 git config --global core.longpaths true 附地址https://github.com/Strider-CD/st ...
- R in a nutshell（连载）
data(cars)dim(cars)names(cars)summary(cars)plot(cars, xlab = "Speed (mph)", ylab = "S ...
- 标题编辑 AndroidTagGroup
地址: https://github.com/2dxgujun/AndroidTagGroup