struck有关与集合有关

what a fucking day
this morning , though I really wanted to take history class which is the last lesson of this term, but I was too sleepy to wake up.
After a silent math class, I went to library to type C code.
that made me want to go die.
hhhhhh

• 射击问题struck

Erin最近很喜欢玩射击游戏。她刚考完了C语言的期末考试，感觉很溜，于是又来到了射击娱乐场放松一下。和上次一样，先从老板那租了一把步枪和装有N发子弹的弹夹。在射击的过程中，Erin每次都有两种选择：从弹夹中取出一颗子弹上膛，或者打一发子弹出去。注意：所有的子弹都从枪口上膛。Erin感觉这有点像C语言课程中的“栈”的特点。因此在打完了这N发子弹之后，她想验证一下这些子弹打出来的顺序是不是真的满足“栈”的特性。假设N颗子弹的编号为1,2,…,N。子弹从弹夹中取出的顺序也是从1到N。给定一个子弹被打出的顺序，你可以帮Erin验证其是否满足“栈”的打出顺序吗？

可能有多个测试输入，第一行给出总共的测试输入的个数T，和每个测试输入的子弹数N。(0 < T < 20, 0 < N < 10)

每个测试输入只有一行：用空格隔开的N个数，表示子弹打出的编号顺序。

输出YES或者NO表示判断结果

例如

INPUT：

2 4

4 3 2 1

4 2 3 1

OUTPUT:

YES

NO
一个数之后的数必须降序排列

#include<stdio.h>int main() {    int t, n;    scanf("%d %d", &t, &n);    while (t--) {        int s[15] = {};        int temp;  // 存放s[i]之后可能存在的第一个比s[i]小的        int flag1;  // 标识s[i]之后是否存在比s[i]小的        int flag2;  // 标识出栈序列是否非法        int i, j, k;        for (i = 1; i <= n; i++)            scanf("%d", &s[i]);        for (i = 1; i < n; i++) {            flag1 = 0;            flag2 = 0;            for (j = i + 1; j <= n; j++)  // 寻找s[i]之后第一个比s[i]小的                if (s[j] < s[i]) {  // s[i]之后存在比s[i]小的                    temp = s[j];  // s[j]是s[i]之后第一个比s[i]小的                    flag1 = 1;                    break;                }            if (flag1 == 1)                for (k = j + 1; k <= n; k++)                    if (s[k] < s[i])  // s[j]之后如果还有比s[i]小的数                        if (s[k] >= temp) {  // 出栈序列非法                            flag2 = 1;                            break;                        }                        else  // 都必须小于它之前的一个比s[i]小的数                            temp = s[k];  // 更新temp            if (flag2 == 1) {                printf("No\n");                break;            }        }        if (flag2 == 0)            printf("Yes\n");        }        return 0;}

• 集合有关交集并集
  There are two groups of some numbers（0~50). each group should be input in a set(remove the duplicate numbers).
  And then output the two set, intersection and union of the two set.
  Input format:
  first line: a couple of numbers and end of -1.
  second line: a couple of numbers and end of -1.
  Output format:
  first line：Output the first set.
  second line: Output the second set.
  third line: Output the intersection of the two set.
  fourth line：Output the union of the two set.
  All the numbers in the set should be output by ascending oder.
  Each line behind the numbers, there is a space ’ ‘.
  For example:
  [Input]
  1 2 3 6 5 4 1 2 3 -1
  3 2 3 2 1 0 -1
  [Output]
  1 2 3 4 5 6
  0 1 2 3
  1 2 3
  0 1 2 3 4 5 6
  [Input]
  0 -1
  1 -1
  [Output]
  0
  1
  0 1
  这是我自己写的代码，很低级，也不够优化，但是比较好理解

#include<stdio.h>int sort(int *p, int n);  // 给数组去重复排序的函数void intersection(int *p, int n, int *q, int m);  // 求交集的函数int un(int *p, int n, int *q, int m); // 求并集的函数 // 函数主体int main() {    int a[100] = {0};    int b[100] = {0};    int length1 = 0, length2 = 0;    //给数组a,b赋值    scanf("%d", &a[length1]);    while (a[length1] != -1) {        length1++;        scanf("%d", &a[length1]);    }    scanf("%d", &b[length2]);    while (b[length2] != -1) {        length2++;        scanf("%d", &b[length2]);    }    sort(a, length1);//给数组a, b 去重复排序并输出    sort(b, length2);    intersection(a, length1, b, length2);//求交集    un(a, length1, b, length2);//求并集    return 0;}int sort(int *p, int n) {    int i, j, k;    //冒泡排序升序    for (i = 0; i < n; i++) {        for (j = 0; j < n - i - 1; j++)            if (p[j] > p[j+1]) {                k = p[j];                p[j] = p[j+1];                p[j+1] = k;            }    }    //去重复输出数组    printf("%d ", p[0]);    for (i = 1; i < n; i++) {        if (p[i] == p[i-1])            continue;        printf("%d ", p[i]);    }    printf("\n");}//这个算法不够优化void intersection(int *p, int n, int *q, int m) {    int i, j;    int t = 0, flag = 0;    int temp[3000] = {0}; // 这个数组比较大，但是为了防止溢出必须要开大一点至少2500    //判断两个数组里面有没有一样的元素，有就写入数组temp中    for (i = 0; i < n; i++) {        for (j = 0; j < m; j++) {            if (p[i] == q[j]) {                temp[t] = p[i];                t++;                flag = 1;  //  如果两个没有交集，则flag为0，就会输出'\0',而不是temp[0]            }        }    }    if (flag != 0) {        sort(temp, t);  //  求得新数组再次去重复排序    } else {        printf("\n");    }}int un(int *p, int n, int *q, int m) {    int i, j, k;    int t[101] = {0};//  如果，a的第一个比b的第一个小，数组t中写入a的第一个数，a的下标加1.如果，a的第一个比b的第一个大，数组t中写入b的第一个数，b的下标加1.如果，相等，写入任何一个，ab下标加1.知道 a或b全都比较完，把剩余的数写入t中    for (i = 0, j = 0, k = 0; ; k++) {        if (p[i] == q[j]) {            t[k] = p[i];            i++;            j++;        } else if (p[i] < q[j]) {            t[k] = p[i];            i++;        } else if (p[i] > q[j]) {            t[k] = q[j];            j++;        }        if ((i == n) || (j == m))            break;    }    if ((i == n) && (j < m)) {        for (j; j < m; j++) {            k++;            t[k] = q[j];        }    } else if (i < n && j == m) {        for (i; i < n; i++) {            k++;            t[k] = p[i];        }    }// 对t去重复排序    sort(t, k+1);}

这是大神的代码我来膜拜一下

#include <stdio.h>// set and sort代码int inputSet(int set[]) {    int n, i, j, temp;    int count = 0;    while (scanf("%d", &n)) {        if (n == -1)            break;        for (i = 0; i < count; ++i) {            if (n == set[i])                break;        }        if (i == count) {            set[count] = n;            count++;        }    }    for (i = 0; i < count - 1; ++i)  //  选择排序    for (j = i + 1; j < count; ++j)        if (set[i] > set[j]) {            temp = set[i];            set[i] = set[j];            set[j] = temp;        }    return count;}// 输出交集void outputITS(int a[], int count_a, int b[], int count_b) {    int i, j;    for (i = 0; i < count_a; ++i) {        for (j = 0; j < count_b; ++j) {            if (a[i] == b[j])                printf("%d ", a[i]);        }    }    printf("\n");}// 输出并集void outputUNI(int a[], int count_a, int b[], int count_b) {    int i, j;    i = j = 0;    while (i != count_a && j != count_b) {        if (a[i] < b[j]) {            printf("%d ", a[i]);            i++;        } else if (a[i] == b[j]) {            printf("%d ", a[i]);            i++;            j++;        } else if (a[i] > b[j]) {            printf("%d ", b[j]);            j++;        }    }    if (i != count_a) {        for (; i < count_a; ++i)            printf("%d ", a[i]);    }    if (j != count_b) {        for (; j < count_b; ++j)            printf("%d ", b[j]);    }    printf("\n");}// 输出集合void output(int a[], int c) {    int i;    for (i = 0; i < c; ++i)        printf("%d ", a[i]);    printf("\n");}int main() {    int a[50];    int b[50];    int count_a, count_b;    count_a = inputSet(a);    count_b = inputSet(b);    output(a, count_a);    output(b, count_b);    outputITS(a, count_a, b, count_b);    outputUNI(a, count_a, b, count_b);    return 0;