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.
Input
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).
Output
For each test, write 11 integer which denotes the number of ways of drawing.
Sample Input
2 2 6
1 1 0
50000 50000 1000000000Sample Output
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;
}
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.
Input
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).
Output
For each test, write 11 integer which denotes the number of ways of drawing.
Sample Input
2 2 6
1 1 0
50000 50000 1000000000Sample Output
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;
}
