2.1 基本计数方法

例题1 uva 11538 http://acm.hust.edu.cn/vjudge/problem/28978

在m*n棋盘上放置 黑白两个皇后,相互攻击的方法数.

解法,正着考虑, 同行,同列, 斜着的情况加在一起, 组合书计算一下, 黑白不同 所以是 P(n,2).

对斜着的角的情况暴力算也过. 1760ms

 1 //#define txtout
 2 //#define debug
 3 #include<bits/stdc++.h>
 4 #define mt(a,b) memset(a,b,sizeof(a))
 5 using namespace std;
 6 typedef long long LL;
 7 const double pi=acos(-1.0);
 8 const double eps=1e-8;
 9 const int inf=0x3f3f3f3f;
10 const int M=1e5+10;
11 int n,m;
12 LL getRow(int r,int c){
13     return 1LL*c*(c-1)*r;
14 }
15 LL getFirst(int r){
16     LL sum=0;
17     for(int i=2;i<=r;i++){
18         sum+=1LL*i*(i-1);
19     }
20     return sum;
21 }
22 LL get(int r,int c){
23     int small=min(r,c);
24     int sub=max(r,c)-small+1;
25     return getRow(sub,small)+2*getFirst(small-1);
26 }
27 LL solve(){
28     return getRow(n,m)+getRow(m,n)+get(n,m)+get(m,n);
29 }
30 int main(){
31     #ifdef txtout
32     freopen("in.txt","r",stdin);
33     freopen("out.txt","w",stdout);
34     #endif // txtout
35     while(~scanf("%d%d",&n,&m),n|m){
36         printf("%lld\n",solve());
37     }
38     return 0;
39 }

对斜着的角的预处理, 快非常多. 0ms

 1 //#define txtout
 2 //#define debug
 3 #include<bits/stdc++.h>
 4 #define mt(a,b) memset(a,b,sizeof(a))
 5 using namespace std;
 6 typedef long long LL;
 7 const double pi=acos(-1.0);
 8 const double eps=1e-8;
 9 const int inf=0x3f3f3f3f;
10 const int M=1e6+10;
11 int n,m;
12 LL sum[M];
13 void init(){
14     sum[0]=0;
15     sum[1]=0;
16     for(int i=2;i<M;i++){
17         sum[i]=sum[i-1]+1LL*i*(i-1);
18     }
19 }
20 LL getRow(int r,int c){
21     return 1LL*c*(c-1)*r;
22 }
23 LL getFirst(int r){
24     return sum[r];
25 }
26 LL get(int r,int c){
27     int small=min(r,c);
28     int sub=max(r,c)-small+1;
29     return getRow(sub,small)+2*getFirst(small-1);
30 }
31 LL solve(){
32     return getRow(n,m)+getRow(m,n)+get(n,m)+get(m,n);
33 }
34 int main(){
35     #ifdef txtout
36     freopen("in.txt","r",stdin);
37     freopen("out.txt","w",stdout);
38     #endif // txtout
39     init();
40     while(~scanf("%d%d",&n,&m),n|m){
41         printf("%lld\n",solve());
42     }
43     return 0;
44 }

end