1084 Broken Keyboard (20)（20 分）
On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen.
Now given a string that you are supposed to type, and the string that you actually type out, please list those keys which are for sure worn out.
Input Specification:
Each input file contains one test case. For each case, the 1st line contains the original string, and the 2nd line contains the typed-out string. Each string contains no more than 80 characters which are either English letters [A-Z] (case insensitive), digital numbers [0-9], or "_" (representing the space). It is guaranteed that both strings are non-empty.
Output Specification:
For each test case, print in one line the keys that are worn out, in the order of being detected. The English letters must be capitalized. Each worn out key must be printed once only. It is guaranteed that there is at least one worn out key.
Sample Input:
7_This_is_a_test
_hs_s_a_es
Sample Output:
7TI#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&-(x))
const int maxn = 100 + 10;
typedef long long ll;
char s1[maxn], s2[maxn];
int hashTable[maxn], out[maxn];
//散列如下：
//A-Z - 10-35;
//0-9 - 0-9;
//_ - 36
void init() {
memset(hashTable, 0, sizeof(hashTable));
memset(out, 0, sizeof(out));
}
int change(char ch) {
int id = -1;
if (ch >= '0' && ch <= '9') { id = ch - '0'; }
else if (ch == '_') { id = 36; }
else if (ch >= 'a' && ch <= 'z') { id = ch - 32 - 65 + 10; }
else if (ch >= 'A' && ch <= 'Z') { id = ch - 65 + 10; }
return id;
}
int main() {
init();
scanf("%s", s1);
scanf("%s", s2);
int len1 = strlen(s1);
int len2 = strlen(s2);
for (int i = 0; i < len2; i++) {
int id = change(s2[i]);
if (id != -1) { hashTable[id] = 1; }
}
for (int i = 0; i < len1; i++) {
int id = change(s1[i]);
if (s1[i] >= '0' && s1[i] <= '9') { if (hashTable[id] == 0 && out[id] == 0) { printf("%c", s1[i]); out[id] = 1; } }
else if (s1[i] == '_') { if (hashTable[id] == 0 && out[id] == 0) { printf("_"); out[id] = 1; } }
else if (s1[i] >= 'a' && s1[i] <= 'z') { if (hashTable[id] == 0 && out[id] == 0) { printf("%c", (char)(s1[i] - 32)); out[id] = 1; } }
else if (s1[i] >= 'A' && s1[i] <= 'Z') { if (hashTable[id] == 0 && out[id] == 0) { printf("%c", s1[i]); out[id] = 1; } }
}
return 0;
}
1092 To Buy or Not to Buy (20)（20 分）
Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence Eva must check whether a string in the shop contains all the beads she needs. She now comes to you for help: if the answer is "Yes", please tell her the number of extra beads she has to buy; or if the answer is "No", please tell her the number of beads missing from the string.
For the sake of simplicity, let's use the characters in the ranges [0-9], [a-z], and [A-Z] to represent the colors. For example, the 3rd string in Figure 1 is the one that Eva would like to make. Then the 1st string is okay since it contains all the necessary beads with 8 extra ones; yet the 2nd one is not since there is no black bead and one less red bead.
\ Figure 1
Input Specification:
Each input file contains one test case. Each case gives in two lines the strings of no more than 1000 beads which belong to the shop owner and Eva, respectively.
Output Specification:
For each test case, print your answer in one line. If the answer is "Yes", then also output the number of extra beads Eva has to buy; or if the answer is "No", then also output the number of beads missing from the string. There must be exactly 1 space between the answer and the number.
Sample Input 1:
ppRYYGrrYBR2258
YrR8RrY
Sample Output 1:
Yes 8
Sample Input 2:
ppRYYGrrYB225
YrR8RrY
Sample Output 1:
No 2#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
int hashTable[maxn];
int extra = 0, miss = 0;
char s1[maxn], s2[maxn];
int change(char ch) {
if (ch >= '0' && ch <= '9') { return ch - '0'; }
else if (ch >= 'a' && ch <= 'z') { return ch - 'a' + 10; }
else if (ch >= 'A' && ch <= 'Z') { return ch - 'A' + 36; }
}
void init() {
memset(hashTable, 0, sizeof(hashTable));
extra = miss = 0;
}
int main()
{
char ch;
init();
scanf("%s", s1);
scanf("%s", s2);
int len1 = strlen(s1);
int len2 = strlen(s2);
for (int i = 0; i < len1; i++) {
int id = change(s1[i]);
hashTable[id]++;
}
for (int i = 0; i < len2; i++) {
int id = change(s2[i]);
hashTable[id]--;
if (hashTable[id] < 0) {
miss++;
}
}
extra = len1 - len2;//如果能够构成第二个字符串的话，那么这个长度就是就是多余的珠子的数量；
if (miss > 0) printf("No %d\n", miss);
else printf("Yes %d\n", len1 - len2);
return 0;
}
STL版
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
map<char, int> mp1;
int miss, extra;
void init() {
miss = extra = 0;
mp1.clear();
}
int main()
{
init();
string s1, s2;
cin >> s1;
cin >> s2;
int len1 = s1.length(), len2 = s2.length();
for (int i = 0; i < len1; i++) {
mp1[s1[i]]++;
}
for (int i = 0; i < len2; i++) {
mp1[s2[i]]--;
if (mp1[s2[i]] < 0) {
miss++;
}
}
if (miss > 0) printf("No %d\n", miss);
else printf("Yes %d\n", len1 - len2);
return 0;
}
1050 String Subtraction (20)（20 分）
Given two strings S~1~ and S~2~, S = S~1~ - S~2~ is defined to be the remaining string after taking all the characters in S~2~ from S~1~. Your task is simply to calculate S~1~ - S~2~ for any given strings. However, it might not be that simple to do it fast.
Input Specification:
Each input file contains one test case. Each case consists of two lines which gives S~1~ and S~2~, respectively. The string lengths of both strings are no more than 10^4^. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.
Output Specification:
For each test case, print S~1~ - S~2~ in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.#include <bits/stdc++.h>
using namespace std;
map<char, int> mp;
void init() {
mp.clear();
}
int main() {
init();
string s1, s2;
getline(cin, s1);
getline(cin, s2);
int len1 = s1.length(), len2 = s2.length();
for (int i = 0; i < len2; i++) mp[s2[i]] = 1;
for (int i = 0; i < len1; i++) {
if (!mp.count(s1[i])) {
cout << s1[i];
}
}
cout << endl;
return 0;
}
1048 Find Coins (25)（25 分）
Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 10^5^ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10^5^, the total number of coins) and M(<=10^3^, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the two face values V~1~ and V~2~ (separated by a space) such that V~1~ + V~2~ = M and V~1~ <= V~2~. If such a solution is not unique, output the one with the smallest V~1~. If there is no solution, output "No Solution" instead.
Sample Input 1:
8 15
1 2 8 7 2 4 11 15
Sample Output 1:
4 11
Sample Input 2:
7 14
1 8 7 2 4 11 15
Sample Output 2:
No Solution#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&-(x))
const int maxn = 1000 + 10;
typedef long long ll;
int n, m;
int hashTable[maxn], a;
void init() {
memset(hashTable, 0, sizeof(hashTable));
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%d", &a);
hashTable[a]++;
}
for (int i = 1; i < m; i++) {
if (hashTable[i] && hashTable[m - i]) {
if (i == m - i && hashTable[i] <= 1) {
continue;
}
printf("%d %d\n", i, m - i);
return 0;
}
}
printf("No Solution\n");
return 0;
}