怎么将123转换成一二三

[zijibingyuan]

如题目

[至善者善之敌]

最简单的办法，查表替换
1 一
2 二
3 三

[bdmh]

对应个大写数组就是了

[赵4老师]

char hznum[10][3]={
    "〇",
    "一",
    "二",
    "三",
    "四",
    "五",
    "六",
    "七",
    "八",
    "九",
};
int n=123,L,i;
char str[10];
char hzstr[40];
L=sprintf(str,"%d",n);
hzstr[0]=0;
for (i=0;i<L;i++) sprintf(hzstr,"%s%s",hzstr,hznum[str[i]-'0']);
printf("%s\n",hzstr);//一二三

[赵4老师]

另外参考这个

#include <stdio.h>
#include <string.h>
char hznum[100];
char hzunit[15][3]={"","十","百","千","万","十","百","千","亿","十","百","千","万","十","百"};
char hzdigit[10][3]={"零","一","二","三","四","五","六","七","八","九"};
char numstr[20];
char *ReadNumByHZ(double v) {
    char *p,*d;
    int u,i;

    sprintf(numstr,"%.15lg",v);
    printf("%s\n",numstr);
    if ('-'==numstr[0]) {
        sprintf(hznum,"不支持读负数");
        return hznum;
    }
    p=strchr(numstr,'e');
    if (NULL!=p) {
        sprintf(hznum,"不支持读太大或太小的数");
        return hznum;
    }
    d=NULL;
    p=strchr(numstr,'.');
    if (NULL!=p) {
        p[0]=0;
        d=p+1;
    }
    u=strlen(numstr);
    hznum[0]=0;
    for (i=0;i<u;i++) {
        if ('0'==numstr[i] && 0!=((u-1-i)%4)) {
            sprintf(hznum,"%s%s",hznum,hzdigit[numstr[i]-'0']);
        } else {
            sprintf(hznum,"%s%s%s",hznum,hzdigit[numstr[i]-'0'],hzunit[u-1-i]);
        }
    }
    printf("%s\n",hznum);
    u=strlen(hznum);
    for (i=0;i<u;i+=2) {
        if (0==strncmp(hznum+i,"零零",4)) {//多个连续的“零”改成一个“零”
            memmove(hznum+i,hznum+i+2,u-i-2+1);
            i-=2;
            u-=2;
        }
    }
    printf("%s\n",hznum);
    if (0==strcmp(hznum+u-2,"零")) {//末尾“零”去掉
        hznum[u-2]=0;
        u-=2;
    }
    printf("%s\n",hznum);
    for (i=0;i<u;i+=2) {
        if (0==strncmp(hznum+i,"零万",4) || 0==strncmp(hznum+i,"零亿",4)) {//“万”“亿”前是“零”去掉
            memmove(hznum+i,hznum+i+2,u-i-2+1);
            u-=2;
        }
    }
    printf("%s\n",hznum);
    if (NULL!=d) {
        sprintf(hznum,"%s点",hznum);
        i=0;
        while (1) {
            if (0==d[i]) break;
            sprintf(hznum,"%s%s",hznum,hzdigit[d[i]-'0']);
            i++;
        }
    }
    printf("%s\n",hznum);
    return hznum;
}
void main() {
    printf("\n%s\n",ReadNumByHZ(130040600090.012));
}
//130040600090.012
//一千三百零零亿四千零六十零万零零九十零
//一千三百零亿四千零六十零万零九十零
//一千三百零亿四千零六十零万零九十
//一千三百亿四千零六十万零九十
//一千三百亿四千零六十万零九十点零一二
//
//一千三百亿四千零六十万零九十点零一二

[紫回蓝]

[Quote=引用 4 楼 zhao4zhong1 的回复:]
另外参考这个

C/C++ code
#include <stdio.h>
#include <string.h>
char hznum[100];
char hzunit[15][3]={"","十","百","千","万","十","百","千","亿","十","百","千","万","十","百"};
char hzdigit[10][3]={"零","一","二","三",……
[/Quote]
顶

[勤奋的小游侠]

++[Quote=引用 4 楼 zhao4zhong1 的回复:]
另外参考这个

C/C++ code
#include <stdio.h>
#include <string.h>
char hznum[100];
char hzunit[15][3]={"","十","百","千","万","十","百","千","亿","十","百","千","万","十","百"};
char hzdigit[10][3]={"零","一","二","三",……
[/Quote]

[hongwenjun]

[Quote=引用 4 楼 zhao4zhong1 的回复:]

另外参考这个C/C++ code
#include <stdio.h>
#include <string.h>
char hznum[100];
char hzunit[15][3]={"","十","百","千","万","十","百","千","亿","十","百","千","万","十","百"};
char hzdigit[10][3]={"零","一","二","三","四","五"……
[/Quote]

赵大牛 贮备的代码真多，标记一下啊，有空读

[wyfwx]

[Quote=引用 1 楼 babilife 的回复:]
最简单的办法，查表替换
1 一
2 二
3 三
[/Quote]
++

[shiweifu]

查表。

[鸟辉]

mark..

[蠓虫带着秤砣飞]

解决了。