1不在数列中的数字

给出一个长度为n的数列，包含1到n的数字，输出1到n中不在数列中的数字。

#include <stdio.h>4
int main(void) {
    int i,j,n,l = 0,m;
    scanf("%d",&n);
    int a[n],b[n];
    for(i=0;i<n;i++){
        if(a[i] == '\n')
        {
            break;
        }
        else{
            scanf("%d",&a[i]);
        }
    }
    for(i=1;i<=n;i++){
        int k = 0;
        for(j=0;j<n;j++){
            if(i != a[j]){
                k++;
            }
        }
        if(k==n){
            b[l] =i;
            l++;
        }
    }
    printf("%d\n",l);
    for(m=0;m<l;m++){
        printf("%d\n",b[m]);
    }
    
    return 0;
}

2输出ascii码

#include <stdio.h>

int main(void) {
    char s[] ="I want to take the MS's rocket to live in the MS's ice castle ";
    int n = 0;
    while(s[n] != '\0'){
        n++;
        if(s[n] == ' '){
            continue;
        }
        else{
            printf("%d\t",s[n]);
        }
    }
    printf("Hello World\n");
    return 0;
}

3尽管很勉强，代码还是凑凑活活出来了

#include <stdio.h>//用换行的方式无法解决问题，或许用数组可以
int main(void) {//c语言没有动态数组，所以无法批量存储字符串
    int a,b,c,d;
    int i,j,k,l;
    //scanf("%d",&a);
    scanf("%d\t%d\t%d\t%d",&a,&b,&c,&d);
    char N[] = "******___**___**___******____*____*____*____*____*____**************************____******____*******____*___******____*____*******____*****____************____******___***********____*____*____*____*******___*******___************___******____******";
    char N0[] = "******___**___**___******";
    char N1[] = "____*____*____*____*____*";
    char N2[] = "____*********************";
    char N3[] = "*****____******____******";
    char N4[] = "*____*___******____*____*";
    char N5[] = "******____*****____******";
    char N6[] = "******____******___******";
    char N7[] = "*****____*____*____*____*";
    char N8[] = "******___*******___******";
    char N9[] = "******___******____******";
    for(i=0;i<5;i++){
        for(k = 0;k <= 9;k++){
            if(a == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(b == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(c == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(d == k){
                printf("%c",N[i + 25*k]);
            }
        }
    }
    printf("\n");
    for(i=5;i<10;i++){
        for(k = 0;k <= 9;k++){
            if(a == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(b == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(c == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(d == k){
                printf("%c",N[i + 25*k]);
            }
        }
    }
    printf("\n");
    for(i=10;i<15;i++){
        for(k = 0;k <= 9;k++){
            if(a == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(b == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(c == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(d == k){
                printf("%c",N[i + 25*k]);
            }
        }
    }
    printf("\n");
    for(i=15;i<20;i++){
        for(k = 0;k <= 9;k++){
            if(a == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(b == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(c == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(d == k){
                printf("%c",N[i + 25*k]);
            }
        }
    }
    printf("\n");
    for(i=20;i<25;i++){
        for(k = 0;k <= 9;k++){
            if(a == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(b == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(c == k){
                printf("%c",N[i + 25*k]);
            }
            printf(" ");
            if(d == k){
                printf("%c",N[i + 25*k]);
            }

        }
    }
    printf("\n");
}

4感觉是玩不出新花样了就干脆考math库了，math库要重点看一下

#include <stdio.h>
#include<math.h>
int main(void) {
    long int i,j,n;
    scanf("%ld",&n);
    /*while((i = getchar()) != '\n'){

    }*/
    long int T[n];
    for(i = 0;i< n ;i++){
        scanf("%ld",&T[i]);
    }
    printf("input over");
    for(i = 0;i < n;i++){
        long int m = 0,k = 0,l =0;
        for(j = 1;j <=T[i];j++){//需要先上取整
            l = ceil(log2(j));
            m = m + l;
        }
        //k = log2(m);
        printf("%ld\n",m%(1000000007));
    }
    printf("Hello World\n");
    return 0;
}

5简单的数学问题，可见考场上心理素质也很重要

#include <stdio.h>

int main(void) {
    int n;
    scanf("%d",&n);
    n = n + 1;

    if(n%3 == 1 || n % 3 == 2){
        printf("Brother Chao will get the gold nugget!");
    }
    else{
        printf("PangPang will get the gold nugget!");
    }
    printf("Hello World\n");
    return 0;
}