HDU 5676 ztr loves lucky numbers （模拟）

ztr loves lucky numbers

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121332#problem/I

Description

ztr loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.

One day ztr came across a positive integer n. Help him to find the least super lucky number which is not less than n.

Input

There are T cases

For each cases:

The only line contains a positive integer . This number doesn't have leading zeroes.

Output

For each cases

Output the answer

Sample Input

2

4500

47

Sample Output

4747

47

题意：

定义super number:

有且仅有4和7两个数字，且两个数字出现的次数相同.

给出n，求最小的不小于n的super number

题解：

xjb打了一通大模拟，结果越打越觉得坑；

还好分清细节后过掉了；

模拟思路：考虑当前数位能是否能放4或7(用strcmp比较后续数);

当两数大小确定后，后面的序列应按最小顺序编排；

另解：

1. 直接打表然后二分.
2. 用next_permutation获取可能的组合再比较.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 3100
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

char num[25];
int cnt;

int main(int argc, char const *argv[])
{
    //IN;
    //freopen("out.txt","w",stdout);

//    printf("%d\n", 1000000);
//    for(int i=1; i<=1000000; i++)
//        printf("%d\n", i);

    int t; cin >> t; getchar();
    //int t = 0;
    while(t--)
    {
        LL n=0,cmps=0;
        char c;
        cnt = 0;
        memset(num,0,sizeof(num));
        gets(num);

        for(int i=0; i<strlen(num); i++) {
            if(num[i]<'0' || num[i]>'9') break;
            cnt++;
            n = n*10 + num[i] - '0';
        }
        num[cnt] = 0;

        if(cnt&1) {
            for(int i=1; i<=(cnt+1)/2; i++) printf("4");
            for(int i=1; i<=(cnt+1)/2; i++) printf("7");
            printf("\n");
            continue;
        }

        for(int i=1; i<=cnt/2; i++) cmps = cmps*10 + 7;
        for(int i=1; i<=cnt/2; i++) cmps = cmps*10 + 4;

        if(cmps < n) {
            for(int i=1; i<=(cnt+2)/2; i++) printf("4");
            for(int i=1; i<=(cnt+2)/2; i++) printf("7");
            printf("\n");
            continue;
        }

        char ans[25] = {0};
        int si=cnt/2,qi=cnt/2;
        for(int i=0; i<cnt; i++) {
            if(num[i]<'4') {
                if(si) ans[i] = '4', si--;
                else ans[i] = '7', qi--;
                for(int j=0; j<=i; j++) printf("%c", ans[j]);
                while(si--) printf("4");
                while(qi--) printf("7");
                printf("\n");
                break;
            }
            if(num[i] == '4') {
                char tmp[25] = {0};
                for(int j=0; j<qi; j++) tmp[j] = '7';
                for(int j=qi; j<qi+si-1; j++) tmp[j] = '4';
                if(!si || strcmp(tmp,num+i+1) < 0) {
                    ans[i] = '7'; qi--;
                    for(int j=0; j<=i; j++) printf("%c", ans[j]);
                    while(si--) printf("4");
                    while(qi--) printf("7");
                    printf("\n");
                    break;
                }
                if(strcmp(tmp,num+i+1) == 0) {
                    ans[i] = '4'; si--;
                    for(int j=0; j<=i; j++) printf("%c", ans[j]);
                    printf("%s",tmp);
                    printf("\n");
                    break;
                }
                ans[i] = '4'; si--;
                continue;
            }
            if(num[i] == '7') {
                char tmp[25] = {0};
                for(int j=0; j<qi-1; j++) tmp[j] = '7';
                for(int j=qi-1; j<qi+si-1; j++) tmp[j] = '4';
                if(!qi || strcmp(tmp,num+i+1) < 0) {
                    for(int i=1; i<=(cnt+2)/2; i++) printf("4");
                    for(int i=1; i<=(cnt+2)/2; i++) printf("7");
                    printf("\n");
                    break;
                }
                if(strcmp(tmp,num+i+1) == 0) {
                    ans[i] = '7'; qi--;
                    for(int j=0; j<=i; j++) printf("%c", ans[j]);
                    printf("%s",tmp);
                    printf("\n");
                    break;
                }
                ans[i] = '7'; qi--;
                continue;
            }

            ans[i] = '7'; qi--;
            for(int j=0; j<=i; j++)
                printf("%c", ans[j]);
            while(si--) printf("4");
            while(qi--) printf("7");
            printf("\n");
            break;
        }

        //printf("\n");
    }

    return 0;
}