Timus 1153. Supercomputer 要求根据自然数列的前 N 项和求 N 值。
Problem Author: Eugene Bryzgalov
Problem Source: Ural Collegiate Programming Contest, April 2001, Perm, English Tour
1153. Supercomputer
Time Limit: 2.0 second
Memory Limit: 16 MB
To check the speed of JCN Corporation new supercomputer it was decided to figure out the sum of first N (N<10600)
positive integers. Unfortunately, by the time the calculation was
finished the Chief Programmer forgot the value of N he entered.
Your task is to write the program (for personal computer), which would
determine the value of N by the result calculated on supercomputer.
Note:
JCN Corporation manufactures only reliable computers, and its programmers write only correctly working programs.
Note:
JCN Corporation manufactures only reliable computers, and its programmers write only correctly working programs.
Input
One line containing number M - the result of calculations on supercomputer.
Output
One line, containing N - number, entered by Chef Programmer.
Sample
| input | output |
|---|---|
| 28 | 7 |
Problem Source: Ural Collegiate Programming Contest, April 2001, Perm, English Tour
解答如下：
1 using System;
2 using System.Globalization;
3
4 namespace Skyiv.Ben.Timus
5 {
6 // http://acm.timus.ru/problem.aspx?space=1&num=1153
7 sealed class T1153
8 {
9 static void Main()
10 {
11 BigInteger m = BigInteger.Parse(Console.ReadLine());
12 Console.WriteLine(BigInteger.Sqrt(m + m));
13 }
14 }
15
16 sealed class BigInteger : IComparable<BigInteger>
17 {
18 int[] digits = new int[601];
19
20 public BigInteger(int n)
21 {
22 digits[0] = n;
23 if (digits[0] > 9) Format();
24 }
25
26 public BigInteger(BigInteger x)
27 {
28 Array.Copy(x.digits, digits, digits.Length);
29 }
30
31 public static BigInteger Parse(string s)
32 {
33 BigInteger x = new BigInteger(0);
34 for (int i = s.Length - 1; i >= 0; i--) x.digits[s.Length - 1 - i] = s[i] - '0';
35 return x;
36 }
37
38 public int CompareTo(BigInteger x)
39 {
40 for (int i = digits.Length - 1; i >= 0; i--)
41 {
42 if (digits[i] > x.digits[i]) return 1;
43 else if (digits[i] < x.digits[i]) return -1;
44 }
45 return 0;
46 }
47
48 public static BigInteger Sqrt(BigInteger x)
49 {
50 BigInteger low, high;
51 GetLowAndHigh(x, out low, out high);
52 BigInteger mid = low;
53 int cmp = 0;
54 while (low.CompareTo(high) <= 0)
55 {
56 mid = (low + high) / 2;
57 cmp = (mid * mid).CompareTo(x);
58 if (cmp < 0) low = mid + 1;
59 else if (cmp > 0) high = mid + (-1);
60 else return mid;
61 }
62 if (cmp > 0) mid = mid + (-1);
63 return mid;
64 }
65
66 static void GetLowAndHigh(BigInteger x, out BigInteger low, out BigInteger high)
67 {
68 int xmax = x.digits.Length - 1;
69 while (xmax >= 0 && x.digits[xmax] == 0) xmax--;
70 const int doublePrecison = 15;
71 if (xmax < doublePrecison)
72 {
73 low = high = new BigInteger((int)Math.Sqrt(GetPrefix(x, xmax, doublePrecison)));
74 return;
75 }
76 int zeros = xmax - doublePrecison + 1;
77 if (zeros % 2 != 0) zeros++;
78 string[] ss = Math.Sqrt(GetPrefix(x, xmax, xmax - zeros + 1)).ToString("F8", CultureInfo.InvariantCulture).Split('.');
79 low = new BigInteger(0);
80 zeros /= 2;
81 int j = 1;
82 for (int i = 0; i < ss[0].Length; i++) low.digits[i + zeros] = ss[0][ss[0].Length - 1 - i] - '0';
83 for (int i = 0; i < ss[1].Length - 2 && j <= zeros; i++, j++) low.digits[zeros - j] = ss[1][i] - '0';
84 high = new BigInteger(low);
85 if (++high.digits[zeros - j + 1] > 9) high.Format();
86 }
87
88 static long GetPrefix(BigInteger x, int start, int length)
89 {
90 long v = 0;
91 for (int i = start; i >= 0 && length > 0; i--, length--) v = v * 10 + x.digits[i];
92 return v;
93 }
94
95 public static BigInteger operator +(BigInteger x, int y)
96 {
97 BigInteger z = new BigInteger(x);
98 z.digits[0] += y;
99 if (z.digits[0] > 9 || z.digits[0] < 0) z.Format();
100 return z;
101 }
102
103 public static BigInteger operator +(BigInteger x, BigInteger y)
104 {
105 BigInteger z = new BigInteger(x);
106 for (int i = x.digits.Length - 1; i >= 0; i--) z.digits[i] = x.digits[i] + y.digits[i];
107 z.Format();
108 return z;
109 }
110
111 public static BigInteger operator *(BigInteger x, BigInteger y)
112 {
113 BigInteger z = new BigInteger(0);
114 int xmax = x.digits.Length - 1;
115 int ymax = y.digits.Length - 1;
116 while (xmax >= 0 && x.digits[xmax] == 0) xmax--;
117 while (ymax >= 0 && y.digits[ymax] == 0) ymax--;
118 for (int xi = xmax; xi >= 0; xi--)
119 for (int yi = ymax; yi >= 0; yi--)
120 z.digits[xi + yi] += x.digits[xi] * y.digits[yi];
121 z.Format();
122 return z;
123 }
124
125 public static BigInteger operator /(BigInteger x, int y)
126 {
127 BigInteger z = new BigInteger(0);
128 int xmax = x.digits.Length - 1;
129 while (xmax >= 0 && x.digits[xmax] == 0) xmax--;
130 for (int remainder = 0, i = xmax; i >= 0; i--)
131 {
132 int quotient = 10 * remainder + x.digits[i];
133 remainder = quotient % y;
134 z.digits[i] = quotient / y;
135 }
136 return z;
137 }
138
139 void Format()
140 {
141 for (int quotient = 0, i = 0; i < digits.Length; i++)
142 {
143 int numerator = digits[i] + quotient;
144 quotient = numerator / 10;
145 int remainder = numerator % 10;
146 if (remainder < 0)
147 {
148 remainder += 10;
149 quotient--;
150 }
151 digits[i] = remainder;
152 }
153 }
154
155 public override string ToString()
156 {
157 int n = digits.Length - 1;
158 while (n >= 0 && digits[n] == 0) n--;
159 if (n < 0) return "0";
160 char[] cs = new char[n + 1];
161 for (int i = n; i >= 0; i--) cs[i] = (char)(digits[n - i] + '0');
162 return new string(cs);
163 }
164 }
165 }
2 using System.Globalization;
3
4 namespace Skyiv.Ben.Timus
5 {
6 // http://acm.timus.ru/problem.aspx?space=1&num=1153
7 sealed class T1153
8 {
9 static void Main()
10 {
11 BigInteger m = BigInteger.Parse(Console.ReadLine());
12 Console.WriteLine(BigInteger.Sqrt(m + m));
13 }
14 }
15
16 sealed class BigInteger : IComparable<BigInteger>
17 {
18 int[] digits = new int[601];
19
20 public BigInteger(int n)
21 {
22 digits[0] = n;
23 if (digits[0] > 9) Format();
24 }
25
26 public BigInteger(BigInteger x)
27 {
28 Array.Copy(x.digits, digits, digits.Length);
29 }
30
31 public static BigInteger Parse(string s)
32 {
33 BigInteger x = new BigInteger(0);
34 for (int i = s.Length - 1; i >= 0; i--) x.digits[s.Length - 1 - i] = s[i] - '0';
35 return x;
36 }
37
38 public int CompareTo(BigInteger x)
39 {
40 for (int i = digits.Length - 1; i >= 0; i--)
41 {
42 if (digits[i] > x.digits[i]) return 1;
43 else if (digits[i] < x.digits[i]) return -1;
44 }
45 return 0;
46 }
47
48 public static BigInteger Sqrt(BigInteger x)
49 {
50 BigInteger low, high;
51 GetLowAndHigh(x, out low, out high);
52 BigInteger mid = low;
53 int cmp = 0;
54 while (low.CompareTo(high) <= 0)
55 {
56 mid = (low + high) / 2;
57 cmp = (mid * mid).CompareTo(x);
58 if (cmp < 0) low = mid + 1;
59 else if (cmp > 0) high = mid + (-1);
60 else return mid;
61 }
62 if (cmp > 0) mid = mid + (-1);
63 return mid;
64 }
65
66 static void GetLowAndHigh(BigInteger x, out BigInteger low, out BigInteger high)
67 {
68 int xmax = x.digits.Length - 1;
69 while (xmax >= 0 && x.digits[xmax] == 0) xmax--;
70 const int doublePrecison = 15;
71 if (xmax < doublePrecison)
72 {
73 low = high = new BigInteger((int)Math.Sqrt(GetPrefix(x, xmax, doublePrecison)));
74 return;
75 }
76 int zeros = xmax - doublePrecison + 1;
77 if (zeros % 2 != 0) zeros++;
78 string[] ss = Math.Sqrt(GetPrefix(x, xmax, xmax - zeros + 1)).ToString("F8", CultureInfo.InvariantCulture).Split('.');
79 low = new BigInteger(0);
80 zeros /= 2;
81 int j = 1;
82 for (int i = 0; i < ss[0].Length; i++) low.digits[i + zeros] = ss[0][ss[0].Length - 1 - i] - '0';
83 for (int i = 0; i < ss[1].Length - 2 && j <= zeros; i++, j++) low.digits[zeros - j] = ss[1][i] - '0';
84 high = new BigInteger(low);
85 if (++high.digits[zeros - j + 1] > 9) high.Format();
86 }
87
88 static long GetPrefix(BigInteger x, int start, int length)
89 {
90 long v = 0;
91 for (int i = start; i >= 0 && length > 0; i--, length--) v = v * 10 + x.digits[i];
92 return v;
93 }
94
95 public static BigInteger operator +(BigInteger x, int y)
96 {
97 BigInteger z = new BigInteger(x);
98 z.digits[0] += y;
99 if (z.digits[0] > 9 || z.digits[0] < 0) z.Format();
100 return z;
101 }
102
103 public static BigInteger operator +(BigInteger x, BigInteger y)
104 {
105 BigInteger z = new BigInteger(x);
106 for (int i = x.digits.Length - 1; i >= 0; i--) z.digits[i] = x.digits[i] + y.digits[i];
107 z.Format();
108 return z;
109 }
110
111 public static BigInteger operator *(BigInteger x, BigInteger y)
112 {
113 BigInteger z = new BigInteger(0);
114 int xmax = x.digits.Length - 1;
115 int ymax = y.digits.Length - 1;
116 while (xmax >= 0 && x.digits[xmax] == 0) xmax--;
117 while (ymax >= 0 && y.digits[ymax] == 0) ymax--;
118 for (int xi = xmax; xi >= 0; xi--)
119 for (int yi = ymax; yi >= 0; yi--)
120 z.digits[xi + yi] += x.digits[xi] * y.digits[yi];
121 z.Format();
122 return z;
123 }
124
125 public static BigInteger operator /(BigInteger x, int y)
126 {
127 BigInteger z = new BigInteger(0);
128 int xmax = x.digits.Length - 1;
129 while (xmax >= 0 && x.digits[xmax] == 0) xmax--;
130 for (int remainder = 0, i = xmax; i >= 0; i--)
131 {
132 int quotient = 10 * remainder + x.digits[i];
133 remainder = quotient % y;
134 z.digits[i] = quotient / y;
135 }
136 return z;
137 }
138
139 void Format()
140 {
141 for (int quotient = 0, i = 0; i < digits.Length; i++)
142 {
143 int numerator = digits[i] + quotient;
144 quotient = numerator / 10;
145 int remainder = numerator % 10;
146 if (remainder < 0)
147 {
148 remainder += 10;
149 quotient--;
150 }
151 digits[i] = remainder;
152 }
153 }
154
155 public override string ToString()
156 {
157 int n = digits.Length - 1;
158 while (n >= 0 && digits[n] == 0) n--;
159 if (n < 0) return "0";
160 char[] cs = new char[n + 1];
161 for (int i = n; i >= 0; i--) cs[i] = (char)(digits[n - i] + '0');
162 return new string(cs);
163 }
164 }
165 }
我们知道，自然数列的前 N 项和 M = N * ( N + 1 ) / 2 ≈ N2 / 2。所以 N = [ √ 2 * M ]。这里 [x] 表示对 x 进行下取整。
这个程序的关键就是求 BigInteger 的平方根。本程序中第 66 到 86 行的 GetLowAndHigh() 方法估算出平方根的范围，然后在第 48 到 64 行的 Sqrt() 方法中用二分查找法找出所求的平方根。
