CF1355C Count Triangles
作者:互联网
思路:
枚举z,对于每个可能的z,计算有多少种可行的x,y组合。
实现:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 5 ll sum(ll l, ll r) 6 { 7 return (l + r) * (r - l + 1) / 2; 8 } 9 10 ll cal(ll z, ll l, ll r) 11 { 12 ll x = min(l, r), y = max(l, r); 13 ll tmp = 0; 14 if (y >= z - 1) tmp = sum(z - x, z - 1); 15 else 16 { 17 tmp = y * min(z - y, x); 18 if (x > z - y) tmp += sum(z - x, y - 1); 19 } 20 return l * r - tmp; 21 } 22 23 ll work(ll z, ll a, ll b, ll c, ll d) 24 { 25 return cal(z, b, d) - cal(z, b, c - 1) - cal(z, a - 1, d) + cal(z, a - 1, c - 1); 26 } 27 28 int main() 29 { 30 ll a, b, c, d; 31 while (cin >> a >> b >> c >> d) 32 { 33 ll res = 0; 34 for (ll z = c; z <= d; z++) res += work(z, a, b, b, c); 35 cout << res << endl; 36 } 37 return 0; 38 }
标签:Count,tmp,return,ll,long,cal,CF1355C,sum,Triangles 来源: https://www.cnblogs.com/wangyiming/p/14772109.html