Initial import of code

LICENSE

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+This is free and unencumbered software released into the public domain.
+
+Anyone is free to copy, modify, publish, use, compile, sell, or
+distribute this software, either in source code form or as a compiled
+binary, for any purpose, commercial or non-commercial, and by any
+means.
+
+In jurisdictions that recognize copyright laws, the author or authors
+of this software dedicate any and all copyright interest in the
+software to the public domain. We make this dedication for the benefit
+of the public at large and to the detriment of our heirs and
+successors. We intend this dedication to be an overt act of
+relinquishment in perpetuity of all present and future rights to this
+software under copyright law.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
+EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
+MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
+IN NO EVENT SHALL THE AUTHORS BE LIABLE FOR ANY CLAIM, DAMAGES OR
+OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE,
+ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR
+OTHER DEALINGS IN THE SOFTWARE.
+
+For more information, please refer to <http://unlicense.org>
+

README.md

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+SPOJ
+====
+
+This repository contains solutions to almost all problems I have solved on
+SPOJ. They're intended as a last resort in case you've tried very hard to solve
+a problem but you just haven't been able to figure it out. When I was trying to
+solve <a href="http://www.spoj.com/problems/GSS2/">GSS2</a>, I mostly found
+vague hints on online forums, which weren't helpful.  I wasn't able to solve
+the problem until after I looked at some code. So obviously the code helps, and
+there can be good reasons to just look at code rather than being stuck
+indefinitely.
+
+The copyright notice (LICENSE) applies to all code except for a few algorithms
+copied from the Stanford ACM team notebook, which are indicated as such. There
+is technically no license for that code that I'm aware of, but it's unlikely
+you'll get into any trouble for using it or redistributing it freely. A recent
+version of the Stanford ACM team notebook can be found <a
+href="http://stanford.edu/~liszt90/acm/notebook.html">here</a>.
+
+There is a comment at the beginning of each file that gives the date on which
+the solution was submitted. Caution: Many of the older solutions have bad
+coding style and poor mastery of C++.
+
+For a few problems I have multiple solutions, usually because after my first
+successful submission I discovered there was a much better way to solve the
+problem. Those are indicated as e.g. niceday-1.cpp and niceday-2.cpp. There is
+no possible ambiguity as SPOJ problem names aren't allowed to contain dashes
+(as far as I know, anyway).

aba12d.cpp

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+// 2014-09-22
+#include <vector>
+#include <algorithm>
+#include <cstdio>
+using namespace std;
+bool is_knumber[1000001];
+int knumber_count[1000001];
+bool is_prime(int x) {
+    if (x < 2) return false;
+    for (int i = 2; i*i <= x; i++) {
+        if (x%i == 0) return false;
+    }
+    return true;
+}
+int main() {
+    is_knumber[2] = true;
+    for (int i = 2; i <= 1000; i++) {
+        if (is_prime(i)) {
+            int pwr = i*i;
+            int sod = 1 + i + i*i;
+            while (pwr <= 1000000) {
+                if (is_prime(sod)) {
+                    is_knumber[pwr] = true;
+                }
+                pwr *= i;
+                sod += pwr;
+            }
+        }
+    }
+    knumber_count[0] = 0;
+    for (int i = 1; i <= 1000000; i++) {
+        knumber_count[i] = knumber_count[i-1] + int(is_knumber[i]);
+    }
+    int T; scanf("%d", &T);
+    while (T--) {
+        int A, B; scanf("%d %d", &A, &B);
+        printf("%d\n", knumber_count[B] - knumber_count[A-1]);
+    }
+    return 0;
+}

abcd.cpp

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+// 2014-05-02
+#include <cstdio>
+char s[100001];
+char t[100001];
+char other(char a, char b, char c) {
+    for (char d = 'A';; d++) {
+        if (d != a && d != b && d != c) return d;
+    }
+}
+int main() {
+    int N; scanf("%d\n", &N);
+    gets(s);
+    for (int i = 0; i < N; i++) {
+        t[2*i] = other(s[2*i], s[2*i+1], i == 0 ? 'A' : t[2*i-1]);
+        t[2*i+1] = other(t[2*i], s[2*i], s[2*i+1]);
+    }
+    t[2*N] = 0;
+    puts(t);
+}

abcdef.cpp

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+// 2014-04-25
+#include <cstdio>
+#include <algorithm>
+#include <vector>
+using namespace std;
+int a[100];
+int main() {
+    int N;
+    scanf("%d", &N);
+    for (int i = 0; i < N; i++) {
+        scanf("%d", a+i);
+    }
+    vector<int> s1, s2;
+    for (int i = 0; i < N; i++) {
+        for (int j = 0; j < N; j++) {
+            for (int k = 0; k < N; k++) {
+                s1.push_back(a[i]*a[j]+a[k]);
+                if (a[i] != 0) {
+                    s2.push_back(a[i]*(a[j]+a[k]));
+                }
+            }
+        }
+    }
+    sort(s1.begin(), s1.end());
+    sort(s2.begin(), s2.end());
+    long long res = 0; int i = 0, j = 0;
+    while (i < s1.size() && j < s2.size()) {
+        if (s1[i] < s2[j]) {
+            i++;
+        } else if (s1[i] > s2[j]) {
+            j++;
+        } else {
+            int k, m;
+            for (k = i; k < s1.size() && s1[k] == s1[i]; k++);
+            for (m = j; m < s2.size() && s2[m] == s2[j]; m++);
+            res += (k-i)*(long long)(m-j); i = k; j = m;
+        }
+    }
+    printf("%lld\n", res);
+}

abcpath.cpp

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+// 2014-05-01
+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+using namespace std;
+const int dx[8] = {1, 1, 1, 0, 0, -1, -1, -1};
+const int dy[8] = {1, 0, -1, 1, -1, 1, 0, -1};
+int main() {
+    for (int cs = 1;; cs++) {
+        int R, C; cin >> R >> C; if (R==0) return 0;
+        vector<string> V(R);
+        vector<vector<int> > dp(R, vector<int>(C, -1e9));
+        for (int i = 0; i < R; i++) {
+            cin >> V[i];
+        }
+        int res = 0;
+        for (char c = 'A'; c <= 'Z'; c++) {
+            for (int i = 0; i < R; i++) {
+                for (int j = 0; j < C; j++) {
+                    if (V[i][j] != c) continue;
+                    if (c == 'A') dp[i][j] = 1;
+                    for (int k = 0; k < 8; k++) {
+                        int r = i + dx[k], c = j + dy[k];
+                        if (r >= 0 && c >= 0 && r < R && c < C &&
+                            V[r][c] == V[i][j] - 1) {
+                            dp[i][j] = max(dp[i][j], dp[r][c] + 1);
+                        }
+                    }
+                    res = max(res, dp[i][j]);
+                }
+            }
+        }
+        cout << "Case " << cs << ": " << res << endl;
+    }
+}

absp1.cpp

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+// 2014-10-07
+#include <cstdio>
+int a[10000];
+int main() {
+    int T; scanf("%d", &T);
+    while (T--) {
+        int N; scanf("%d", &N);
+        long long res = 0;
+        long long sum = 0;
+        for (int i = 0; i < N; i++) {
+            scanf("%d", a + i);
+            if (i > 0) {
+                res += i*(long long)a[i] - sum;
+            }
+            sum += a[i];
+        }
+        printf("%lld\n", res);
+    }
+    return 0;
+}

absys.cpp

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+// 2008-06-16
+#include <iostream>
+#include <cstring>
+#define MAX(a,b) ((a)>(b)?(a):(b))
+#define MIN(a,b) ((a)<(b)?(a):(b))
+#define MAXS 10000
+using namespace std;
+struct bignum
+{
+	char* digits;
+	int length;
+	bignum(int size)
+	{
+		length=1;
+		digits=new char[size];
+		digits[0]=0;
+	}
+	bignum(int size,const bignum& x)     //Copy constructor
+	{
+		length=x.length;
+		digits=new char[size];
+		memcpy(digits,x.digits,x.length);
+	}
+	bignum(int size,int x)
+	{
+		length=0;
+		digits=new char[size];
+		while (x>0)
+		{
+			digits[length++]=x%10;
+			x/=10;
+		}
+		if (length==0) {digits[0]=0; length=1;}
+	}
+	~bignum()
+	{
+		delete digits;
+	}
+	void add(bignum& x)
+	{
+		int l=MAX(length,x.length);
+		int d;
+		int carry=0;
+		x.digits[x.length]=0;
+		memset(digits+length,0,l-length+1);
+		for (d=0; d<=l; d++)
+		{
+			int sum=carry;
+			if (d<length)
+				sum+=digits[d];
+			if (d<x.length)
+				sum+=x.digits[d];
+			digits[d]=sum;
+			if (digits[d]>=10) {digits[d]-=10;carry=1;} else carry=0;
+		}
+		length=l;
+		if (digits[length]) length++;
+		if (length==0) {digits[0]=0; length=1;}
+	}
+	void sub(bignum& x)
+	{
+		int d;
+		int borrow=0;
+		digits[length]=0;
+		for (d=0; d<=length; d++)
+		{
+			digits[d]-=borrow;
+			if (d<x.length)
+				digits[d]-=x.digits[d];
+			if (digits[d]<0) {digits[d]+=10;borrow=1;} else borrow=0;
+		}
+		while (length>0&&!digits[length-1]) length--;
+		if (length==0) length=1;
+	}
+	void print()
+	{
+		int i;
+		for (i=length-1; i>=0; i--)
+			putchar(digits[i]+48);
+	}
+	void operator=(const bignum& y)
+	{
+		length=y.length;
+		memcpy(digits,y.digits,y.length);
+	}
+	void operator=(const char* s)
+	{
+		length=strlen(s);
+		int i=length;
+		while (i--)
+			digits[i]=s[length-i-1]-48;
+	}
+};
+int main()
+{
+	int i,j,T;
+	scanf("%d",&T);
+	char s1[MAXS],s2[MAXS],s3[MAXS];
+	bignum n1(MAXS),n2(MAXS),n3(MAXS);
+	for (j=0; j<T; j++)
+	{
+		scanf("%s + %s = %s",&s1,&s2,&s3);
+		//one of the numbers has a "machula"
+		for (i=0; i<strlen(s1); i++)
+			if (s1[i]=='m')
+			{
+				n2=s2;
+				n3=s3;
+				n1=n3;
+				n1.sub(n2);
+			}
+		for (i=0; i<strlen(s2); i++)
+			if (s2[i]=='m')
+			{
+				n1=s1;
+				n3=s3;
+				n2=n3;
+				n2.sub(n1);
+			}
+		for (i=0; i<strlen(s3); i++)
+			if (s3[i]=='m')
+			{
+				n1=s1;
+				n2=s2;
+				n3=n1;
+				n3.add(n2);
+			}
+		n1.print(); printf(" + "); n2.print(); printf(" = "); n3.print(); printf("\n");
+	}
+	return 0;
+}