1.[a,b]的奇数之和

[1,2k-1]的奇数之和：k^2

111.png

((b+1)/2)((b+1)/2)-(a/2)(a/2)

2.[a,b]的偶数之和：

[1,2k]的偶数和

k+k^2

：(b/2-(a-1)/2)+(b/2)(b/2)-((a-1)/2)((a-1)/2)

3.八进制十进制小数

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#include<iostream>
#include<cstring>
using namespace std;
const int N = 10010;
char d[N] ;  //八进制数
int ans[N] ;  //十进制数
int main(){
    while(cin>>d){
        memset(ans,0,sizeof(ans));  //每组数据都必须先初始化ans[] 
        int d2;
        int len = strlen(d);       //记录小数的位数 
        int t = 0 ;
        for(int i = len - 1 ; i > 1 ; i--){
            d2 = d[i] - '0' ;     //d2来接收小数的每一位的数 
            int k = 0 , j = 0 ;
            while(j<t || d2){    //此循环内语句为数组模拟除法计算 
                d2 = d2*10 + ans[j++]; 
                ans[k++] = d2/8 ;
                d2 %= 8 ;
            }
            t = k;     //记录最后的得到的位数 
        }
        cout<<d<<" [8] = 0." ;
        for(int i = 0 ; i < t ; i++)
            cout<<ans[i];
        cout<<" [10]"<<endl;
    }
    return 0;
}

十进制转R进制

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

char digits[] = "0123456789ABCDEF";
#define N 32
char ans[N + 2];

int main()
{
    int n, r;
    while (~scanf ("%d%d", &n, &r)) {
        if(r==0)break;
        if (n < 0) {printf("-"); n = -n;}
        int i = 0;
        while (n) ans[i++] = digits[n % r], n /= r;
        if (i == 0) ans[i++] = '0';

        /* 输出结果 */
        while (--i >= 0) putchar(ans[i]);
        putchar('\n');
    }

    return 0;
}

A转B

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#include <iostream>
#include <string>
#include <ctype.h>
using namespace std;
string convert(int a, string& s, int b)
{
    long val, dcount, digit;
    char result[72], c;
    string ans;
 
    val = 0;
    for(int i=0; i<(int)s.size(); i++) {
        if(isdigit(s[i]))
            val = val * a + s[i] - '0';
        else
            val = val * a + toupper(s[i]) - 'A' + 10;
    }
    dcount = 0;
    while(val) {
        digit = val % b;
        val /= b;
        result[dcount++] = ((digit >= 10) ? 'A' - 10 : '0') + digit;
    }
    if(dcount == 0) {
        result[dcount++] = '0';
        result[dcount] = '\0';
    } else
        result[dcount] = '\0';
    for(int i=0, j=dcount-1; i<j; i++, j--) {
        c = result[i];
        result[i] = result[j];
        result[j] = c;
    }
    ans = result;
    return ans;
}
 
int main()
{
    int a, b;
    string s;
    cin >> a >> s >> b;
    cout << convert(a, s, b) << endl;
    return 0;
}

2转16位数爆longlong

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/* Bailian2798 2进制转化为16进制 */
 
#include <stdio.h>
#include <string.h>
 
#define N 10000
#define N2 4
char s[N + N2 + 1];
 
char convert[] = "0123456789ABCDEF";
 
int main(void)
{
    int n, len, digits, i, k;
 
    scanf("%d", &n);
    getchar();
    while(n--) {
        gets(s + N2);
 
        s[0] = s[1] = s[2] = s[3] = '0';
        len = strlen(s + N2);               /* 2进制位数 */
        digits = (len + N2 - 1) / N2;   /* 16进制位数 */
        len += N2 - 1;
        for(i=1, k=len; i<=digits; i++) {
            s[k--] = convert[(s[len - 3] - '0') * 8 + (s[len - 2] - '0') * 4 + (s[len - 1] - '0') * 2 + (s[len] - '0')];
            len -= N2;
        }
 
        printf("%s\n", &s[k + 1]);
    }
 
    return 0;
}

4.大小写转换

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大写变小写、小写变大写 : 字符 ^= 32;

大写变小写、小写变小写 : 字符 |= 32;

小写变大写、大写变大写 : 字符 &= -33;
a-z：97-122
A-Z：65-90
0-9：48-57
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    int t = x;
    int value = 0;
    while (x / 10){
        value = 10 * value + x % 10;
        x /=  10;
    }
    if(value*10 + x == t) return 1;
    else return 0;
}判断回文数字
  /*幻方*/ #include<iostream>
using namespace std;
 int n,a[40][40],x,y;
int main(){
    cin>>n;
    x=1,y=(n+1)/2;
    for(int i=1;i<=n*n;i++){
        a[x][y]=i;
        if(!a[(x-2+n)%n+1][y%n+1]) x=(x-2+n)%n+1,y=y%n+1;
        else x=x%n+1;
    }
for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++){cout<<a[i][j]<<' ';}
        cout<<endl;}
}