NOIP2001提高组复赛B 数的划分

题目链接：https://ac.nowcoder.com/acm/contest/249/B

题目大意：

　　略

分析1（记忆化搜索）：

　　方法为减而治之，把n划分成k份的答案就相当于每次把n分成a，b两个数，再把a分成k-1份，然后把每次a分成k-1份的答案相加即可。注意点是每轮分出来的b要不大于上一轮分出来的b。

代码如下：

 #include <bits/stdc++.h>
 using namespace std;

 #define rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))

 #define pii pair<int,int>
 #define piii pair<pair<int,int>,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 const int maxN = 1e5 + ;

 int n, k;
 // f[i][j][k]表示数i分成j分的分法总数，k为限制条件，每种分法每份的值不能超过k，用来排除重复
 // f[i][j][k] = f[i-1][j-1][1] + f[i-2][j-1][2] + ……+ f[i-min(k, i-1)][j-1][min(k, i-1)]
 int f[][][];

 int solve(int x, int y, int z){
     int ret = ;
     if(x < y) return ;
     if(y == ) return x <= z ?  : ;
     if(f[x][y][z]) return f[x][y][z];

     For(i, , x-) {
         if(x-i > z) continue;
         ret += solve(i, y-, x-i);
     }
     f[x][y][z] = ret;
     return ret;
 }

 int main(){
     scanf("%d%d", &n, &k);
     printf("%d\n", solve(n, k, ));
     return ;
 }

分析2（DP）：

　　见代码内注释

代码如下：

 #include <bits/stdc++.h>
 using namespace std;

 #define rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))

 #define pii pair<int,int>
 #define piii pair<pair<int,int>,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 const int maxN = 1e5 + ;

 int n, k;
 // f[i][j]表示数i分成j份的分法总数
 /*
     当i < j时，很明显没法分，所以f[i][j] = 0；
     当i == j时，只有一种分法，所以f[i][j] = 1；
     当i > j时，考虑从小到大分，第1个如果分1，那么f[i][j] = f[i-1][j-1]；
   第1个如果分大于1的数，可以对所有j份都减一，然后再分，即 f[i][j] = f[i-j][j]；
   根据加法原则，f[i][j] = f[i-1][j-1] + f[i-j][j]；
 */
 int f[][]; 

 int main(){
     scanf("%d%d", &n, &k);
     For(i, , n) f[i][] = ; // 无论什么数，分成一份都只有一种
     For(i, , k)
         For(j, , n)
             if(j >= i) f[j][i] = f[j-][i-] + f[j-i][i];

     printf("%d\n", f[n][k]);
     return ;
 }