LASTDIG

LASTDIG - The last digit

#include <iostream>

#define ll long long int

using namespace std;

ll expo(ll a, ll b, int n)

{

ll d = 1;

while(b)

{

if((b & 1))

{

d = (d * a) % n;

}

b = b >> 1;

a = (a * a) % n;

}

return d;

}

int main()

{

int t;

ll a, b;

cin >> t;

while(t--)

{

cin >> a >> b;

cout << expo(a, b, 10) << endl;

}

return 0;

}

Comments

GCJ101BB - Picking Up Chicks

Problem Link /* explanation lets solve the problem only for 2 chicken. s[i] = speed of chicken i pos[i] = position of chicken i if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not. if s[i] < s[i - 1] then there is a chance that i can slow down i - 1. lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B. for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec. and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec. at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision. if i is pos[i] < B-T then i can reach B before T sec and it will not cause any problem . problem may occur if pos[i - 1]< (B -T)*2 as it can me

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++)

SPOJ

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