标签：超时 方向 std iostream else 推出 最大 lse cstring
frog has a piece of paper divided into nn rows and mm columns. Today, she would like to draw a rectangle whose perimeter is not greater than kk.
There are 88 (out of 99) ways when n=m=2,k=6n=m=2,k=6
Find the number of ways of drawing.
The input consists of multiple tests. For each test:
The first line contains 33 integer n,m,kn,m,k (1≤n,m≤5?104,0≤k≤1091≤n,m≤5?104,0≤k≤109).
For each test, write 11 integer which denotes the number of ways of drawing.
2 2 6
1 1 0
50000 50000 1000000000 8
0
1562562500625000000
解析：枚举矩形的长h，然后它在h的方向上就有n-h+1种放法，同时宽的最大值w即为k/2 - h，对于每个0~w的宽度wi，它在w方向上的放法有m-wi+1种，求和即为所求方案数
我推出了数学公式
n*m中a*b的种数：（n-a+1）*(m-b+1)+(m-a+1)*(n-b+1)
这样仍会超时：a不变，b变化，推出一个公式。
1^2+2^2+3^2+……+n^2=n*(n+1)*(2*t+1)/6;
#include<iostream> #include<cmath> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #define ll long long using namespace std; int main() { ll n,m,k; while(~scanf("%lld %lld %lld",&n,&m,&k)) { ll s=0; ll temp; if(k<6&&k>=4) s=n*m; else if(k<4) s=0; else { if(n>m) { temp=n; n=m; m=temp; } for(ll a=1;a<=n;a++) { ll b=min(k/2-a,m); if(b>=a) { s+=(n-a+1)*((m-a+1)+(m-b+1))*(b-a+1)/2; } b=min(k/2-a,n); if(b>=a) { s+=(m-a+1)*((n-a+1)+(n-b+1))*(b-a+1)/2; } } ll t=min(k/4,n); t=min(t,m); ll ss=0; ss+=(t*(t+1)*(2*t+1))/6-(t+1)*t*(n+m+2)/2+(n*m+1+n+m)*t;//中间有重复的情况，a=b的算了两次 s=s-ss; } printf("%lld\n",s); } return 0; }
也有更简单的思路：
#include <bits/stdc++.h> using namespace std; int main(){ #ifdef sxk freopen("in.txt", "r", stdin); #endif // sxk long long n, m, k; while(cin>>n>>m>>k){ k /= 2; long long ans = 0; for(int h=1; h<=n; h++){ int w = k - h; if(w <= 0) break; if(w > m) w = m; ans += (n - h + 1) * (m + m-w+1)*w/2; //对于每个h，在h方向上n-h+1种，在w方向上枚举wi求和为(m + m-w+1) * w / 2种 } cout<<ans<<endl; } return 0; }
标签：超时 方向 std iostream else 推出 最大 lse cstring
原文地址：https://www.cnblogs.com/caiyishuai/p/8955121.html
