POJ1102-模拟

来源：互联网发布：手机淘宝掌柜热卖截图编辑：程序博客网时间：2024/04/28 07:21

题目：题目链接

题意：给出一个数字，按照要求的宽度大小输出LD式的数字表示

分析：模拟，把n是1的情况先一个数字一个数字搞出来。然后再根据n放大。蛋疼的模拟......

代码：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int QuickMod(int  a,int b,int n){    int r = 1;    while(b)    {        if(b&1)            r = (r*a)%n;        a = (a*a)%n;        b >>= 1;    }    return r;}char nums[10][5][3] ={    {{' ', '-', ' '}, {'|', ' ', '|'}, {' ', ' ', ' '}, {'|', ' ', '|'}, {' ', '-', ' '}},    {{' ', ' ', ' '}, {' ', ' ', '|'}, {' ', ' ', ' '}, {' ', ' ', '|'}, {' ', ' ', ' '}},    {{' ', '-', ' '}, {' ', ' ', '|'}, {' ', '-', ' '}, {'|', ' ', ' '}, {' ', '-', ' '}},    {{' ', '-', ' '}, {' ', ' ', '|'}, {' ', '-', ' '}, {' ', ' ', '|'}, {' ', '-', ' '}},    {{' ', ' ', ' '}, {'|', ' ', '|'}, {' ', '-', ' '}, {' ', ' ', '|'}, {' ', ' ', ' '}},    {{' ', '-', ' '}, {'|', ' ', ' '}, {' ', '-', ' '}, {' ', ' ', '|'}, {' ', '-', ' '}},    {{' ', '-', ' '}, {'|', ' ', ' '}, {' ', '-', ' '}, {'|', ' ', '|'}, {' ', '-', ' '}},    {{' ', '-', ' '}, {' ', ' ', '|'}, {' ', ' ', ' '}, {' ', ' ', '|'}, {' ', ' ', ' '}},    {{' ', '-', ' '}, {'|', ' ', '|'}, {' ', '-', ' '}, {'|', ' ', '|'}, {' ', '-', ' '}},    {{' ', '-', ' '}, {'|', ' ', '|'}, {' ', '-', ' '}, {' ', ' ', '|'}, {' ', '-', ' '}}};int main(){    int i, x, jj, kk;    int n, m, digits[20];    while(scanf("%d%d", &n, &m) == 2)    {        if(!n && !m)            break;        i = 0;        if(m == 0) //注意0        {            x = 1;            digits[0] = 0;        }        else        {            //转为单个数字放在数组里            while(m > 0)            {                digits[i] = m % 10;                m = m / 10;                i++;            }            x = i;         }        for(int j = 0; j < 2*n+3; ++j)        {            jj = j;            if(j > 0 && j <= n)                jj = 1;            if(j == n+1)                jj = 2;            if(j > n+1 && j < 2*n+3-1)                jj = 3;            if(j == 2*n+3-1)                jj = 4;            for(i = x-1; i >= 0; --i)            {                for(int k = 0; k < n+2; ++k)                {                    kk = 0;                    if(k > 0 && k < n+1)                        kk = 1;                    if(k == n+1)                        kk = 2;                    printf("%c", nums[digits[i]][jj][kk]);                }                printf(" "); //数字之间要有空格            }            printf("\n");        }        printf("\n"); //输出一个数据后要有空行    }    return 0;}

蛋疼.....