柳婼の blog

给出两个字符串s1和s2，判断 s2里面所有的字母s1都有，有的话返回 true 否则返回 false

#include <iostream>
#include <set>
using namespace std;
int main() {
    string s1, s2;
    cin >> s1 >> s2;
    set<char> t1, t2;
    for(int i = 0; i <= s1.length(); i++) {
        t1.insert(s1[i]);
    }
    for(int i = 0; i <= s2.length(); i++) {
        t2.insert(s2[i]);
    }
    
    set<char>::iterator it1 = t1.begin();
    set<char>::iterator it2 = t2.begin();
    while(it1 != t1.end() && it2 != t2.end()) {
        if(*it1 == *it2) {
            it2++;
        }
        it1++;
    }
    if(it2 != t2.end()) {
        cout << "false";
    } else {
        cout << "true";
    }
    return 0;
}

#include <iostream>

#include <set>

using namespace std;

int main() {

string s1, s2;

cin >> s1 >> s2;

set<char> t1, t2;

for(int i = 0; i <= s1.length(); i++) {

t1.insert(s1[i]);

}

for(int i = 0; i <= s2.length(); i++) {

t2.insert(s2[i]);

}

set<char>::iterator it1 = t1.begin();

set<char>::iterator it2 = t2.begin();

while(it1 != t1.end() && it2 != t2.end()) {

if(*it1 == *it2) {

it2++;

}

it1++;

}

if(it2 != t2.end()) {

cout << "false";

} else {

cout << "true";

}

return 0;

}

#include <iostream>
using namespace std;
int book[26];
int main() {
    string s1, s2;
    cin >> s1 >> s2;
    for(int i = 0; i < s1.length(); i++) {
        book[s1[i] - 'A'] = 1;
    }
    for(int i = 0; i < s2.length(); i++) {
        if(book[s2[i] - 'A'] == 0) {
            cout << "false";
            return 0;
        }
    }
    cout << "true";
    return 0;
}

#include <iostream>

using namespace std;

int book[26];

int main() {

string s1, s2;

cin >> s1 >> s2;

for(int i = 0; i < s1.length(); i++) {

book[s1[i] - 'A'] = 1;

}

for(int i = 0; i < s2.length(); i++) {

if(book[s2[i] - 'A'] == 0) {

cout << "false";

return 0;

}

cout << "true";

return 0;

}

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>

double toSum(char*, int*);
int isPrime(int number);
void twertySixPrimeNumber(int *numbers);
double isContain(double d1, double d2);

int main(int arg, char** argvs) {
    char *s1 = "BAZZY";
    char *s2 = "YA";
    int primeNumbers[26] = { 0 };
    twentySixPrimeNumber(primeNumbers);

    double d1 = toSum(s1, primeNumbers);
    double d2 = toSum(s2, primeNumbers);

    double r = isContain(d1, d2);
    printf("s1: %s s2:%s \n", s1, s2);
    if (r) {
        printf("YES");
    } else {
        printf("NO");
    }
}

double isContain(double d1, double d2) {
    double a = d1 / d2;
    double intpart;
    return 0 == modf(a, &intpart) ? 1 : 0;
}

double toSum(char *s1, int *numbers) {
    double sum = 1;
    for(int i = 0; i < strlen(s1) ; i++ ) {
        char c = toupper(s1[i]);
        int temp = c - 65;
        int primeNumber = numbers[temp];
        sum *= primeNumber;
    }
    return sum;
}

void twentySixPrimeNumber(int *numbers) {
    int counter = 26;
    for (int i = 2 ; i < INT_MAX; i++) {
        if (!counter) break;
        if ( isPrime(i) ) {
            numbers[26-counter] = i;
            counter--;
        }
    }
}

int isPrime(int number) {
    if ( 1 == number ||  2 == number || 3 == number) return 1;
    for (int i = 2; i < sqrt(number); i++) {
        if (!(number % i))
          { return 0;}
        else { return 1;}
    }
}

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

#include <limits.h>

double toSum(char*, int*);

int isPrime(int number);

void twertySixPrimeNumber(int *numbers);

double isContain(double d1, double d2);

int main(int arg, char** argvs) {

char *s1 = "BAZZY";

char *s2 = "YA";

int primeNumbers[26] = { 0 };

twentySixPrimeNumber(primeNumbers);

double d1 = toSum(s1, primeNumbers);

double d2 = toSum(s2, primeNumbers);

double r = isContain(d1, d2);

printf("s1: %s s2:%s \n", s1, s2);

if (r) {

printf("YES");

} else {

printf("NO");

}

double isContain(double d1, double d2) {

double a = d1 / d2;

double intpart;

return 0 == modf(a, &intpart) ? 1 : 0;

}

double toSum(char *s1, int *numbers) {

double sum = 1;

for(int i = 0; i < strlen(s1) ; i++ ) {

char c = toupper(s1[i]);

int temp = c - 65;

int primeNumber = numbers[temp];

sum *= primeNumber;

}

return sum;

}

void twentySixPrimeNumber(int *numbers) {

int counter = 26;

for (int i = 2 ; i < INT_MAX; i++) {

if (!counter) break;

if ( isPrime(i) ) {

numbers[26-counter] = i;

counter--;

}

int isPrime(int number) {

if ( 1 == number || 2 == number || 3 == number) return 1;

for (int i = 2; i < sqrt(number); i++) {

if (!(number % i))

{ return 0;}

else { return 1;}

}

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>

int judge(int vector, int vectorS2);
int toVector(char*);

int main(int arg, char** argvs)
{
    char *s1 = "ACBSDEREJASLKDASD";
    char *s2 = "jbd";

    int v1 = toVector(s1);
    int v2 = toVector(s2);
    int v3 = v1 & v2;

    if ( 1 == judge(v3, v2) ) {
        printf("%s is in the %s ", s2, s1);
    } else {
        printf("%s was not in the %s ", s2, s1);
    }
}

int judge(int vector, int vectorS2) {
    if (abs((vector - vectorS2)) < 0.1) {
        return 1;
    } else {
        return vector == 0 ? -1 : 0;
    }
}

int toVector(char *s1) {
    int vector = 0;
    for(int i = 0; i < strlen(s1) ; i++ ) {
        char c = toupper(s1[i]);
        int temp = c - 65;
        vector |= (1 << temp);
    }
    return vector;
}

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <ctype.h>

int judge(int vector, int vectorS2);

int toVector(char*);

int main(int arg, char** argvs)

{

char *s1 = "ACBSDEREJASLKDASD";

char *s2 = "jbd";

int v1 = toVector(s1);

int v2 = toVector(s2);

int v3 = v1 & v2;

if ( 1 == judge(v3, v2) ) {

printf("%s is in the %s ", s2, s1);

} else {

printf("%s was not in the %s ", s2, s1);

}

int judge(int vector, int vectorS2) {

if (abs((vector - vectorS2)) < 0.1) {

return 1;

} else {

return vector == 0 ? -1 : 0;

}

int toVector(char *s1) {

int vector = 0;

for(int i = 0; i < strlen(s1) ; i++ ) {

char c = toupper(s1[i]);

int temp = c - 65;

vector |= (1 << temp);

}

return vector;

}

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.

思路：
设甲乙一人一次为一轮。进行了很多轮之后，让甲选的时候，如果是1，2，3那就可以通过。如果是4一定不能赢，所以如果是5，6，7可以想办法分别取1，2，3让乙来一定不能赢，所以5，6，7甲是可以赢的。
如果是8，则无论甲选任何个数，都能让乙来面临7，6，5这些必赢的选项所以甲一定输。
如果是9，10，11，则甲可以通过让乙来面临8来一定输。
如果是12，则甲无论选取任何个数，都能让乙面临9，10，11这样的一定可以赢的数字，所以12让甲必定输。
以此类推发现规律，在4或者4的倍数的时候，甲无论怎样一定输。
所以就简单一句：return n % 4

class Solution {
public:
    bool canWinNim(int n) {
        return n % 4;
    }
};

class Solution {

public:

bool canWinNim(int n) {

return n % 4;

}

};

中午闲聊时候的一道题(^0_0^)

LeetCode 292. Nim Game

xcode 中不小心删除 stroryboard 的恢复办法

iOS开发：remove reference与move to trash的区别

报错： Consecutive statements on a line must be separated by ‘;’

解决办法：Transport security has blocked a cleartext HTTP (http://) resource load since it is insecure. Temporary exceptions can be configured via your app’s Info.plist file.