```
#include<bits/stdc++.h>
using namespace std;
long long s[30][30][30];
long long x,y,z,t=0;
long long f(long long a,long long b,long long c){
if(a<=0||b<=0||c<=0){
return 1;
}else if(a>20||b>20||c>20){
if(t==0){
s[20][20][20]=f(20,20,20);
t=1;
}
return s[20][20][20];
}
if(s[a][b][c]>0){
return s[a][b][c];
}
if(a<b&&b<c){
s[a][b][c]=f(a,b,c-1)+f(a,b-1,c-1)-f(a,b-1,c);
return s[a][b][c];
}else{
s[a][b][c]=f(a-1,b,c)+f(a-1,b-1,c)+f(a-1,b,c-1)-f(a-1,b-1,c-1);
return s[a][b][c];
}
}
int main(){
int s=0;
int jjj=1;
while(jjj=1){
cin>>x>>y>>z;
if(x==-1&&y==-1&&z==-1) return 0;
s=f(x,y,z);
cout<<"w("<<x<<", "<<y<<", "<<z<<") = "<<s<<endl;
}
return 0;
}
```
by changzang @ 2023-09-30 11:26:22
```
// 2023/9/29 15:37
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define INF 2147483646
int ans[21][21][21];
int w(int a, int b, int c) {
if (a <= 0 || b <= 0 || c <= 0) {
return 1;
}
if (ans[a][b][c] != 0) {
return ans[a][b][c];
}
if (a < b && b < c) {
int k = w(a, b, c - 1) + w(a, b - 1, c - 1) - w(a, b - 1, c);
ans[a][b][c] = k;
return k;
}
int k = w(a - 1, b, c) + w(a - 1, b - 1, c) + w(a - 1, b, c - 1) -
w(a - 1, b - 1, c - 1);
ans[a][b][c] = k;
return k;
}
int main() {
// for (int i = 0; i <= 20; i++) {
// for (int j = 0; j <= 20; j++) {
// for (int k = 0; k <= 20; k++) {
// ans[i][j][k] = 0;
// }
// }
// } 1. 全局变量默认初始化为 0
ll a, b, c;
scanf("%lld %lld %lld", &a, &b, &c);
// while (a > -1 || b > -1 || c > -1) {
// 2. 保证输入的数在 [−9223372036854775808,9223372036854775807] 之间，并且是整数。
// 你这个就不循环了
while (1) {
if (a <= 0 || b <= 0 || c <= 0) {
printf("w(%lld, %lld, %lld) = 1\n", a, b, c);
} else {
cout << "w(" << a << ", " << b << ", " << c << ")";
if (a > 20) {// 3. 注意：例如 w(30,-1,0) 又满足条件 1 又满足条件 2，请按照最上面的条件来算，答案为 1。
// 如果有一个 > 20，修改其他的 -1，你的解法就相当于先满足条件 2 再满足条件 1 了
a = 20;
b = 20;
c = 20;
} else if (b > 20) {
a = 20;
b = 20;
c = 20;
} else if (c > 20) {
a = 20;
b = 20;
c = 20;
}
cout << " = " << w(a, b, c) << endl;// 4. 如果不需要清空缓存区，且输出次数较多，如
// 保证不包括 $-1, -1, -1$ 的输入行数 $T$ 满足 $1 \leq T \leq 10 ^ 5$。
// 建议使用 "\n"，如 cout << " = " << w(a, b, c) << "\n";
}
cin >> a >> b >> c;
if (a == -1 && b == -1 && c == -1) break;
}
system("pause");
return 0;
}
/*----------------------------------------------------*/
/*测试
输入
-1 21 21
-1 -2 -1234567
2147483648 1 1
-1 -1 -1
输出
w(-1, 21, 21) = 1
w(-1, -2, -1234567) = 1
w(2147483648, 1, 1) = 1048576
*/
/*----------------------------------------------------*/
#include <bits/stdc++.h> // 示例代码
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int maxn = 25;
ll p[maxn][maxn][maxn];
ll w(ll a, ll b, ll c) {
if (a <= 0 || b <= 0 || c <= 0) return 1;
if (p[a][b][c]) return p[a][b][c];// 放在递归 w 之前
// if (a > 20 || b > 20 || c > 20) return p[a][b][c] = w(20, 20, 20);
if (a < b && b < c) return p[a][b][c] = w(a, b, c - 1) + w(a, b - 1, c - 1)
- w(a, b - 1, c);
return p[a][b][c] = w(a - 1, b, c) + w(a - 1, b - 1, c) + w(a - 1, b, c -
1)
- w(a - 1, b - 1, c - 1);
}
int main() {
ll a, b, c, d, e, f;
while (scanf("%lld %lld %lld", &a, &b, &c) == 3) {
if (a == -1 && b == -1 && c == -1) break;
d = a; e = b; f = c;
if (a <= 0 || b <= 0 || c <= 0) {
printf("w(%lld, %lld, %lld) = 1\n", a, b, c);
} else {
if (a > 20 || b > 20 || c > 20) d = e = f = 20;
printf("w(%lld, %lld, %lld) = %lld\n", a, b, c, w(d, e, f));
}
}
system("pause");
return 0;
}
```
by Sqj147 @ 2023-10-06 17:28:23