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：
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define rep(i,n) for (int i = 0; i < (n); ++i) 5 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 6 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 7 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 8 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 9 10 #define pr(x) cout << #x << " = " << x << " " 11 #define prln(x) cout << #x << " = " << x << endl 12 13 #define ALL(x) x.begin(),x.end() 14 #define INS(x) inserter(x,x.begin()) 15 16 #define ms0(a) memset(a,0,sizeof(a)) 17 #define msI(a) memset(a,inf,sizeof(a)) 18 19 #define pii pair<int,int> 20 #define piii pair<pair<int,int>,int> 21 #define mp make_pair 22 #define pb push_back 23 #define fi first 24 #define se second 25 26 inline int gc(){ 27 static const int BUF = 1e7; 28 static char buf[BUF], *bg = buf + BUF, *ed = bg; 29 30 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 31 return *bg++; 32 } 33 34 inline int ri(){ 35 int x = 0, f = 1, c = gc(); 36 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 37 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 38 return x*f; 39 } 40 41 typedef long long LL; 42 const int maxN = 1e5 + 7; 43 44 int n, k; 45 // 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 46 // 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)] 47 int f[201][7][202]; 48 49 int solve(int x, int y, int z){ 50 int ret = 0; 51 if(x < y) return 0; 52 if(y == 1) return x <= z ? 1 : 0; 53 if(f[x][y][z]) return f[x][y][z]; 54 55 For(i, 1, x-1) { 56 if(x-i > z) continue; 57 ret += solve(i, y-1, x-i); 58 } 59 f[x][y][z] = ret; 60 return ret; 61 } 62 63 int main(){ 64 scanf("%d%d", &n, &k); 65 printf("%d\n", solve(n, k, 201)); 66 return 0; 67 }
分析2（DP）：
见代码内注释
代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define rep(i,n) for (int i = 0; i < (n); ++i) 5 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 6 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 7 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 8 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 9 10 #define pr(x) cout << #x << " = " << x << " " 11 #define prln(x) cout << #x << " = " << x << endl 12 13 #define ALL(x) x.begin(),x.end() 14 #define INS(x) inserter(x,x.begin()) 15 16 #define ms0(a) memset(a,0,sizeof(a)) 17 #define msI(a) memset(a,inf,sizeof(a)) 18 19 #define pii pair<int,int> 20 #define piii pair<pair<int,int>,int> 21 #define mp make_pair 22 #define pb push_back 23 #define fi first 24 #define se second 25 26 inline int gc(){ 27 static const int BUF = 1e7; 28 static char buf[BUF], *bg = buf + BUF, *ed = bg; 29 30 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 31 return *bg++; 32 } 33 34 inline int ri(){ 35 int x = 0, f = 1, c = gc(); 36 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 37 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 38 return x*f; 39 } 40 41 typedef long long LL; 42 const int maxN = 1e5 + 7; 43 44 int n, k; 45 // f[i][j]表示数i分成j份的分法总数 46 /* 47 当i < j时，很明显没法分，所以f[i][j] = 0； 48 当i == j时，只有一种分法，所以f[i][j] = 1； 49 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]； 50 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]； 51 根据加法原则，f[i][j] = f[i-1][j-1] + f[i-j][j]； 52 */ 53 int f[201][7]; 54 55 int main(){ 56 scanf("%d%d", &n, &k); 57 For(i, 1, n) f[i][1] = 1; // 无论什么数，分成一份都只有一种 58 For(i, 2, k) 59 For(j, 2, n) 60 if(j >= i) f[j][i] = f[j-1][i-1] + f[j-i][i]; 61 62 printf("%d\n", f[n][k]); 63 return 0; 64 } 65