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 ;
}
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