2014
02-14

# Consecutive Digits

As a recruiting ploy, Google once posted billboards in Harvard Square and in the Silicon Valley area just stating “{first 10-digit prime found in consecutive digits of e}.com”. In other words, find that 10-digit sequence and then connect to the web site― and find out that Google is trying to hire people who can solve a particular kind of problem. Not to be outdone, Gaggle (a loosy-goosy fuzzy logic search firm), has devised its own recruiting problem. Consider the base 7 expansion of a rational number. For example, the first few digits of the base 7 expansion of 1/510 = 0.12541…7,33/410 = 11.15151…7, and 6/4910 = 0.06000…7, From this expansion, find the digits in a particular range of positions to the right of the "decimal" point.

The input file begins with a line containing a single integer specifying the number of problem sets in the file. Each problem set is specified by four base 10 numbers on a single line, n d b e, where n and d are the numerator and
denominator of the rational number and 0 ≤ n ≤ 5,000 and 1 ≤ d ≤ 5,000. b and e are the beginning and ending positions for the desired range of digits, with 0 ≤ b,e ≤ 250 and 0 ≤ (e-b) ≤ 20. Note that 0 is the position immediately to the right of the decimal point.

The input file begins with a line containing a single integer specifying the number of problem sets in the file. Each problem set is specified by four base 10 numbers on a single line, n d b e, where n and d are the numerator and
denominator of the rational number and 0 ≤ n ≤ 5,000 and 1 ≤ d ≤ 5,000. b and e are the beginning and ending positions for the desired range of digits, with 0 ≤ b,e ≤ 250 and 0 ≤ (e-b) ≤ 20. Note that 0 is the position immediately to the right of the decimal point.

4
1 5 0 0
6 49 1 3
33 4 2 7
511 977 122 126

Problem set 1: 1 / 5, base 7 digits 0 through 0: 1
Problem set 2: 6 / 49, base 7 digits 1 through 3: 600
Problem set 3: 33 / 4, base 7 digits 2 through 7: 151515
Problem set 4: 511 / 977, base 7 digits 122 through 126: 12425

int main()
{
int cases, x = 1;
cin>>cases;
while (cases--) {
int a, b, c, d;
cin>>a>>b>>c>>d;
printf("Problem set %d: %d / %d, base 7 digits %d through %d: ", x++, a, b, c, d);
int i;
a %= b;
for (i = 0; i <= d; ++i) {
a *= 7;
if (i >= c) {
printf("%d", a / b);
}
a %= b;
}
printf("\n");
}
return 0;
}

1. “再把所有不和该节点相邻的节点着相同的颜色”，程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的

2. 第二个方法挺不错。NewHead代表新的头节点，通过递归找到最后一个节点之后，就把这个节点赋给NewHead，然后一直返回返回，中途这个值是没有变化的，一边返回一边把相应的指针方向颠倒，最后结束时返回新的头节点到主函数。

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