USACO 2.2 Runaround Numbers

Problem 57: Runaround Numbers

Runaround numbers are integers with unique digits, none of which is zero (e.g., 81362) that also have an interesting property, exemplified by this demonstration:

• If you start at the left digit (8 in our number) and count that number of digits to the right (wrapping back to the first digit when no digits on the right are available), you'll end up at a new digit (a number which does not end up at a new digit is not a Runaround Number). Consider: 8 1 3 6 2 which cycles through eight digits: 1 3 6 2 8 1 3 6 so the next digit is 6.
• Repeat this cycle (this time for the six counts designed by the `6') and you should end on a new digit: 2 8 1 3 6 2, namely 2.
• Repeat again (two digits this time): 8 1
• Continue again (one digit this time): 3
• One more time: 6 2 8 and you have ended up back where you started, after touching each digit once. If you don't end up back where you started after touching each digit once, your number is not a Runaround number.

Given a number M (that has anywhere from 1 through 9 digits), find and print the next runaround number higher than M, which will always fit into an unsigned long integer for the given test data.

PROGRAM NAME: runround

INPUT FORMAT

A single line with a single integer, M

SAMPLE INPUT (file runround.in)

81361

OUTPUT FORMAT

A single line containing the next runaround number higher than the input value, M.

SAMPLE OUTPUT (file runround.out)

81362

 

类型：数字处理

题意：runround数的判断，runround的定义见题目描述。

思路：1.brute force; 2.runround的定义可知，只包含唯一的1~9的数字，所以可以生成这些数，然后再判定。生成可以用深搜实现。

代码

/*
ID: superbi1
TASK: runround
LANG: C
*/
#include <stdio.h>
#include <string.h>
#include <math.h>

int r[13];
//把数字按位拆分
void div(unsigned int k)
{
int w, p, i;
memset(r, 0, sizeof(r));
w = (int)log10(k);
i = 0;
while (w >= 0) {
p = (int)pow(10, w);
r[++i] = k/p;
k %= p;
w--;
}
}
//判断是否是runround数
int check(unsigned int k)
{
int st = 1, w;
int t = 0, cnt = 0, inc;
int flg[13], i;
memset(flg, 0, sizeof(flg));
w = (int)log10(k)+1;
div(k);
for (i=1; i<=w; i++) {
if (r[i] == 0) return 0;
if (!flg[r[i]]) flg[r[i]] = 1;
else return 0;
}
memset(flg, 0, sizeof(flg));
flg[st] = 1;
cnt++;
while (1) {
inc = r[st];
st = (inc+st)%w;
if (st == 0) st = w;
if (flg[st]) break;
flg[st] = 1;
cnt++;
}
if (cnt == w && st == 1) return 1;
else return 0;
}

int main()
{
FILE *in = fopen("runround.in", "r");
FILE *out = fopen("runround.out", "w");
unsigned int M, i;
fscanf(in, "%u", &M);
for (i=M+1; ; i++) {
if (check(i)) {
fprintf(out, "%u\n", i);
break;
}
}
return 0;
}

 

深搜生成函数：

代码

void
permutation(char *s, int *used, int nd, int md)
{
int i;

if(nd == md) {
s[nd] = '\0';
if(atoi(s) > m && isrunaround(s)) {
fprintf(fout, "%s\n", s);
exit(0);
}
return;
}

for(i=1; i<=9; i++) {
if(!used[i]) {
s[nd] = i+'0';
used[i] = 1;
permutation(s, used, nd+1, md);
used[i] = 0;
}
}
}

 

 

 

转载于:https://www.cnblogs.com/superbin/archive/2010/07/12/1775563.html