SPOJ added summatory problems

Author: sinagarajan

SPOJ/MathFormulas/SumOfCubesUptoN_BEST_PUCMM210.cpp

@@ -0,0 +1,29 @@
+// This is copied from Internet but algorithm is strange 
+
+// This can be used for contests , since I can't remember this algorithm, I didnt use it
+
+#include <cstdio>
+
+typedef long long i64;
+
+const int MOD = 1000000003;
+
+inline i64 mod(i64 a) {
+return a < MOD ? a : a % MOD;
+}
+
+int main() {
+int test, i;
+i64 sum, n, p[6], co[6] = {0, 4, 30, 50, 30, 6}, inv = 441666668;
+scanf("%d", &test);
+while(test--) {
+scanf("%lld", &n);
+for(i = 1, p[0] = 1; i <= 5; i++) p[i] = mod(n * p[i-1]);
+for(i = 1; i <= 5; i++) p[i] = mod(co[i] * p[i]);
+for(i = 1, sum = 0; i <= 5; i++) sum = mod(sum + p[i]);
+sum = mod(sum * inv);
+printf("%lld\n", sum);
+}
+return 0;
+}
+

SPOJ/MathFormulas/SumOfCubesUptoN_PUCMM210.cpp

@@ -0,0 +1,69 @@
+/*
+f(n) is defined as: f(n) = 13+23+33+...+n3, so it is the sum of the cubes of all natural numbers up to n.
+
+In this problem you are about to compute,
+
+f(1) + f(2) + f(3) + ... + f(n)
+Input
+
+The first line is an integer T(1 = T = 100,000), denoting the number of test cases. Then, T test cases follow.
+
+For each test case, there is an integer n(1 = n = 1,000,000) written in one line.
+Output
+
+For each test case output the result of the summatory function described above.
+
+Since this number could be very large, output the answer modulo 1,000,000,003.
+Example
+
+Input:
+3
+2
+10
+3
+
+Output:
+10
+7942
+46
+
+ALGORITHM :
+
+1. Dynamically store all computed values in array 
+2. Access the final value in O(1)
+
+*/
+
+
+
+#include <cstdio>
+
+typedef long long ullong;
+
+const int MOD = 1000000003;
+
+inline ullong mod(ullong input) {
+return input < MOD ? input : input % MOD;
+}
+
+int main() {
+int test, i;
+ullong sum=0, n ,array[1000001], temp;
+array[0]=0;
+scanf("%d", &test);
+
+for(ullong i=1 ; i<1000001; ++i)
+{
+    temp = mod(mod(i*i)*i);
+    sum += temp;
+    array[i] = mod(array[i-1]+sum);
+}
+
+while(test--) 
+{
+scanf("%lld", &n);
+printf("%lld\n",array[n]);
+}
+return 0;
+}
+