LASTDIG2

LASTDIG2 - The last digit re-visited

#include <iostream>

using namespace std;

string s;

int main()

{

int t, i;

long long int a, b, d;

cin >> t;

while(t--)

{

cin >> s;

cin >> b;

i = s.size() - 1;

a = (s[i] - '0');

d = 1;

while(b)

{

if(b & 1)

{

d = (d * a) % 10;

}

a = (a * a) % 10;

b = b >> 1;

}

cout << d << endl;

}

return 0;

}

Comments

GCJ101BB - Picking Up Chicks

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War of XORs- XORIER

Problem Link #include <iostream> using namespace std; int main() { int t, n, odd, even; cin >> t; while(t--) { cin >> n; int i,arr[n],freq[1100001]={0}; long res = 0; odd = even = 0; for(int i = 0; i < n; i++) { cin >> arr[i]; freq[arr[i]]++; } for(int i = 0; i < n; i++) { if(arr[i] & 1) { odd++; } else { even++; } } for(int i = 0; i < n; i++) { if(arr[i] % 2) { res += odd; } else { res += even; } res -= freq[arr[i] ^ 2]; res -= freq[arr[i]]; } cout << res / 2 << endl; } }

MAIN72

MAIN72 - Subset sum #include <iostream> #include <cstring> using namespace std; bool dp[100001][1001]; int arr[1001]; int main() { int t, n; long long int sum; cin >> t; while(t--) { cin >> n; memset(dp, false, sizeof(dp)); sum = 0; for(int i = 0; i < n; i++) { cin >> arr[i]; sum += arr[i]; } for(int i = 0; i < n; i++) dp[0][i] = true; // 0 sum for(int i = 1; i < n; i++) dp[i][0] = false; // sum is i but 0 element for(long int i = 1; i <= sum; i++) { for(int j = 1; j <= n; j ++) { dp[i][j] = dp[i][j - 1]; if(i >= arr[j - 1]) dp[i][j] = dp[i][j] || dp[i - arr[j - 1]][j - 1]; } } long int result = 0; for(int i = 1; i <= sum; i++)

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