洛谷 P1957 口算练习题

 

 

 

 实现代码（原创）：

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

char* itoa(int value, char* str, int radix) {
    static char dig[] =
        "0123456789"
        "abcdefghijklmnopqrstuvwxyz";
    int n = 0, neg = 0;
    unsigned int v;
    char* p, *q;
    char c;
	/* 负数：十进制直接在前面加负号，其它进制用补码 */
    if (radix == 10 && value < 0) {
        v = -value;
        neg = 1;
    }else
    {
        v = value;
    }
    /* 其它进制的负数转为unsigned int类型，即取了补码 */
    do {
        str[n++] = dig[v%radix];
        v /= radix;
    } while (v);
    if (neg)
        str[n++] = '-';
    str[n] = '\0';
	/* 字符串反转 */
    for (p = str, q = p + (n-1); p < q; ++p, --q)
        c = *p, *p = *q, *q = c;
    return str;
}


int main()
{
    int nums[60];
    char str[60][32];
    int n;
    scanf("%d",&n);
    getchar();
    char sym[6];
    char s2;
    int i,j,k;
    int m=0;
    while(n)
    {
        scanf("%s",sym);
        //getchar();
        if(strlen(sym)==1 && (sym[0]=='a'||sym[0]=='b'||sym[0]=='c'))
        {
            char ss1[20],ss2[20],ss3[20];
            int l1,l2,l3;
            int x,y,z;
                s2 = sym[0];
                scanf("%d%d",&x,&y);
                itoa(x,ss1,10);
                itoa(y,ss2,10);
                l1 = strlen(ss1);
                l2 = strlen(ss2);
                if(s2=='a')
                {
                    z = x+y;
                    itoa(z,ss3,10);
                    l3 = strlen(ss3);
                    s2='+';
                }else if(s2=='b')
                {
                    z = x-y;
                    itoa(z,ss3,10);
                    l3 = strlen(ss3);
                    s2='-';
                }else if(s2=='c')
                {
                    z = x*y;
                    itoa(z,ss3,10);
                    l3 = strlen(ss3);
                    s2='*';
                }
                for(i=0;i<l1;i++)
                    str[m][i] = ss1[i];
                str[m][i] = s2;
                i++;
                for(k=0;k<l2;k++,i++)
                    str[m][i] = ss2[k];
                str[m][i] = '=';
                i++;
                for(j=0;j<l3;j++,i++)
                    str[m][i] = ss3[j];
                nums[m]=l1+l2+l3+2;
                m++;

        }else
        {
            char ss2[20],ss3[20];
            int l1,l2,l3;
            int x,y,z;
            x = atoi(sym);
            scanf("%d",&y);

                itoa(y,ss2,10);
                l1 = strlen(sym);
                l2 = strlen(ss2);
                if(s2=='+')
                {
                    z = x+y;
                    itoa(z,ss3,10);
                    l3 = strlen(ss3);
                }else if(s2=='-')
                {
                    z = x-y;
                    itoa(z,ss3,10);
                    l3 = strlen(ss3);
                }else if(s2=='*')
                {
                    z = x*y;
                    itoa(z,ss3,10);
                    l3 = strlen(ss3);
                }
                for(i=0;i<l1;i++)
                    str[m][i] = sym[i];
                str[m][i] = s2;
                i++;
                for(k=0;k<l2;k++,i++)
                    str[m][i] = ss2[k];
                str[m][i] = '=';
                i++;
                for(j=0;j<l3;j++,i++)
                    str[m][i] = ss3[j];
                nums[m]=l1+l2+l3+2;
                m++;

        }
        n--;
    }
    for(i=0;i<m;i++)
    {
        for(j=0;j<nums[i];j++)
            printf("%c",str[i][j]);
        printf("\n");
        printf("%d\n",nums[i]);
    }
    return 0;
}