实验1.1半数集问题 #include using namespace std; int a[1005]={0}; //存放已计算的数据 int b[1005]={0}; int halfset(int n){ int sum = 1; if(a[n]>0) //如果f(n)已经得出,就不必再重复计算 return a[n]; for(int i=1;i<=n/2;i++){ sum += halfset(i); } a[n]=sum; return sum; } int main() { int n = 0,i = 0; cin>>n; while(n!=0){ b[i++] = n; cin>>n; } for(int j=0;j #include #include void mov(int n,char a,char b) { printf("%d %c %c\n",n,a,b); } void dg(int n,char a, char c, char b) { if(n==1) mov(n,a,b); else { dg(n-1,a,b,c); mov(n,a,b); dg(n-1,c,a,b); } } int main() { char a='A',b='B',c='C'; int n; while(scanf("%d",&n)!=EOF) dg(n,a,c,b); return 0; } 实验1.3 整数因子分解问题 #include using namespace std; mapa; int f(int n){ if(n == 1) return 1; if(a[n]) return a[n]; int count=1; for(int i=2;i<=sqrt(n);i++){ if(n%i == 0){ count+=f(i); if(i!=n/i){ count+=f(n/i); } } } a[n] = count; return a[n]; } int main() { int n; while(cin>>n){ cout< #include using namespace std; int a[200][200]; int cost[200]; int city,result; int travel(){ cost[city-2] = a[city-2][city-1]; for(int i = city-3;i>=0;i--){ result = a[i][city-1]; for(int j = i+1;j>city; for(int i = 0;i>a[i][j]; } } result = travel(); cout< using namespace std; const int N = 1e3 + 10; #define Ma_x 99999 #define Mi_x 0 #define min(a,b) ab?a:b int n,w[N],dp[N][N],dq[N][N]; int sum(int i,int t){ int k,s=0,k1; for(k=i;k #include using namespace std; string a,b; int k; int value[10000][10000]; int main(){ cin>>a>>b>>k; int alen=a.size(); int blen=b.size(); for(int i=0;i<=alen;i++){ value[i][0]=k*i; } for(int i=1;i<=blen;i++){ value[0][i]=k*i; } for(int i=1;i<=alen;i++){ for(int j=1;j<=blen;j++){ value[i][j]=min(min(value[i-1][j]+k,value[i][j-1]+k),value[i-1][j-1]+abs(a[i-1]-b[j-1])); } } cout< #include using namespace std; int n; int m[15][15]; int f[25][15][15]; int main() { cin >> n; int x, y, z; while (cin >> x >> y >> z) { if (!x && !y && !z) break; m[x][y] = z; } for (int k = 2; k <= 2 * n; k ++ ) for (int i = 1; i <= n; i ++ ) for (int j = 1; j <= n; j ++ ) { int a = k - i, b = k - j; if (a >= 1 && a <= n && b >= 1 && b <= n) { int t = max(max(f[k - 1][i - 1][j], f[k - 1][i][j]), max(f[k - 1][i][j - 1], f[k - 1][i - 1][j - 1])); if (i == j) f[k][i][j] = t + m[i][a]; else f[k][i][j] = t + m[i][a] + m[j][b]; } } cout << f[2 * n][n][n]; return 0; } 实验2.5双胞胎做题 #include using namespace std; int f[1000][1000]; int a[10000],b[10000]; int n,sum,result=9999; void resulta(){ memset(f, 0, sizeof(f)); for(int i=1;i<=n;i++){ for(int j=0;j<=sum;j++){ if(jt) result=t; } } int main(){ cin>>n; for(int i=1;i<=n;i++){ cin>>a[i]; sum=sum+a[i]; } for(int i=1;i<=n;i++){ cin>>b[i]; } resulta(); cout< using namespace std; typedef long long ll; ll dp[100002],n,m,s[100002]={0}; mapT; struct node{ ll u;ll t; }a[100002]; int main() { cin>>n>>m; for(int i=0;i=1;i--){ if(s[i]==0){dp[i]=dp[i+1];} else{ for(int j=0;j #include #include using namespace std; struct huichang { int s; int f; }; huichang t[10000+10]; int n; bool cmp(huichang a, huichang b) { return a.s < b.s; } int meeting() { int count_chang = 0; int *a = new int[n + 1]; //该数组于保存每个会场的结束时间 for (int i = 0; i < n+1; i++) { a[i] = 0; } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (a[j] <= t[i].s) { a[j] = t[i].f; if (j >= count_chang + 1) count_chang++; break; } } } delete[]a; return count_chang; } int main() { cin >> n; for (int i = 1; i <= n; i++) { cin >> t[i].s >> t[i].f; } sort(t+1,t+n+1,cmp); cout << meeting() << endl; return 0; } 实验3.2磁带最优存储问题 #include #include #include #include using namespace std; struct Tape { int len; //长度 int pr; //概率 double tr; friend bool operator< (Tape a,Tape b) {return a.tr>b.tr;} Tape(int l=0,int p=0):len(l),pr(p),tr(l*p){}; }; priority_queuetape; double minTimeCost(int n,priority_queue&tape,long long sum){ double res=0,acc=0; while(!tape.empty()) { acc+=tape.top().tr/sum;//计算tr累加值 res+=acc; tape.pop(); } return res; } int main() { int n,a,b; long long sum=0; cin>>n; for(int i = 0;i>a>>b; tape.push(Tape(a,b)); sum+=b;//计算概率和 } double res=minTimeCost(n,tape,sum); cout< using namespace std; typedef long long ll; typedef pair PII; const int N = 1e5 + 10; ll n; ll dp[110][110]; int main() { ios::sync_with_stdio(0); cin.tie(0); cin >> n; memset(dp, 0, sizeof dp); dp[1][1] = 1;// i个人分j个宿舍 for (int i = 0; i <= n; i++) { dp[i][0] = 0; dp[i][1] = 1; dp[0][i] = 0; } ll sum = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { if(dp[i][j]==0) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j; } } for (int i = 1; i <= n; i++) { sum += dp[n][i]; } cout << sum; return 0; //17 } 作业2.2烟台基地宿舍分配2 #define _CRT_SECURE_NO_WARNINGS 1 #include using namespace std; typedef long long ll; typedef pair PII; const int N = 1e5 + 10; ll n,m; ll dp[110][110]; int main() { ios::sync_with_stdio(0); cin.tie(0); cin >> n>>m; memset(dp, 0, sizeof dp); dp[1][1] = 1;// i个人分j个宿舍 for (int i = 0; i <= n; i++) { dp[i][0] = 0; dp[i][1] = 1; dp[0][i] = 0; } ll sum = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { if(dp[i][j]==0) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j; } } cout << dp[n][m]; return 0; } 作业3.1金字塔寻宝 #include #include using namespace std; const int N=105; int n,*num,*lastNum,ary1[N],ary2[N]; void dp() { num=ary1,lastNum=ary2; for(int i=1;i<=n;i++){ for(int j=1;j<=i;j++){ cin>>num[j]; num[j]=max(lastNum[j],lastNum[j-1])+num[j]; } swap(num,lastNum); } } int maxValue() { int max=lastNum[1]; for(int i=2;i<=n;i++) if(lastNum[i]>max) max=lastNum[i]; return max; } int main() { cin>>n;dp(); cout< using namespace std; void getnumber(string s); void getresult(); const int N=1010; int f[N][N],num[N][N],a[N]; int d[N][N]; int ans[N]; int cnt; int n,k; int main(){ string s; int c,e; cin>>n>>k; cin>>s; k--; getnumber(s); getresult(); cout< using namespace std; const int N=10000; int servicetime[N]; int line[N]; int main(){ int n,s,sum,num; cin>>n>>s; for(int i=0;i>servicetime[i]; } sort(servicetime,servicetime+n); sum=servicetime[0]+s; num=1; for(int i=0;i using namespace std; const int N=10000; int servicetime[N]; int line[N]; int main(){ int n,s,sum; cin>>n>>s; for(int i=0;i>servicetime[i]; } sort(servicetime,servicetime+n); for(int i=0;i