int main() { int x=1; int i =0; int n=0; int sum=0; while(n<=10) { for(i =1;i<=n;i++) { x*=i; sum=sum+x; } n++; } Return 0; }

C++实现1-n连续整数求和的简易代码解析

i <= n; ++i) { sum += i; } return sum; } int main() { int n; cout << "Enter the value of n: "; cin >> n; // 从用户输入获取n的值 // 调用函数并输出结果 int result = calculateSum(n); cout << ...

C语言实现1至100累加之和的while循环算法

while (i <= 100) { sum += i; // 等价于 sum = sum + i; i++; // 等价于 i = i + 1; } printf("1加到100的总和是：%d\n", sum); return 0; } 这段代码首先包含了stdio.h头文件，它允许我们使用printf...

main()int i=1,sum=0while (i <= 100) sum =sum+1; i++;while (i <= 10) sum =sum+1; i++;print("i=％d，sum=％d",i,sum)

int i = 1, sum = 0; while (i <= 100) { sum += 1; i++; } while (i <= 10) { sum += 1; i++; } printf("i=%d, sum=%d\n", i, sum); return 0; } 这个程序会先将 sum 加一百次，然后再将 sum 加十...

#include <iostream> using namespace std; int Loop(int x) { int sum = 0, n = x; while(n > 0) { sum = sum * 10 + n % 10; n = n / 10; } if(x == sum) return 1; else return 0; } int main() { int a, b, i, n; do{ cin >> a >> b; }while(a >= b); for(i = a; i <= b; i ++) { n = Loop(i); if(i % 2 == 0 && n == 1) cout << i << endl; } return 0; }增加这段代码的范围

if(i % 2 == 0 && n == 1) cout << i << endl; } return 0; } 这样修改后，用户需要输入起始数和结束数，如果起始数大于等于结束数，则会提示用户重新输入。这样就增加了代码的范围，使得用户可以输入任意...

#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include<string.h> #include<ctype.h> int main() { int n; int a[6002] = { 0 }; scanf("%d", &n); int sum= n; int flag = 0; int blag = 0; int count=0; for (int i = 1; i <= n; i++) { a[i] = 1; } int j = 3, k = 2; while (sum > 3) { if (flag == 0) { count = 0; for (int i = 1; i <= n; i++) { count++; if (count % 3 == 0) { a[i] = 2; sum--; } } flag = 1; } if (flag == 1) { count = 0; for (int i = 1; i <= n; i++) { count++; if (count % 2 == 0) { a[i] == 2; sum--; } } flag = 0; } } for (int i = 1; i <= n; i++) { if (a[i] != 0 && a[i] != 2) { if (blag == 0) { printf("%d", i); blag = 1; } else printf(" %d", i); } } return 0; }这个代码啥问题

4. 在第 16 行，变量 sum 的初始值应该为 n，而不是 0，否则会导致程序运行错误。 5. 在第 20 行，应该将数组 a 的前 n 个元素初始化为 1，而不是 0，否则可能会导致程序运行错误。 6. 在第 42 行和...

修改代码 #include<stdio.h> int factorcount(int n){ int count=0; for (int i=1;i<=n;i++){ if(n%i==0){ count++; } } return count; } int main( ) n = int(input()) int n,total=0; for _ in range(n): n = int(input()) total = sum(top_f) { scanf("%d",&n); for (int i=1;i<=n;i++){ total+=factorcount(i); } printf("%d",total); return 0; }

i <= n; i++) { if (n % i == 0) { count++; } } return count; } int main() { int T, n, total; scanf("%d", &T); while (T--) { scanf("%d", &n); total = 0; for (int i = 1; i <= n; i++) { ...

改写成python语言import java.util.LinkedList; import java.util.List; import java.util.Scanner; public class Main { private static final int N = 1000 + 10; private static int[] get_cost(){ int[] cost = new int[N]; boolean[] vis = new boolean[N]; for(int i=1;i<N;i++){ cost[i] = -1; vis[i] = false; } LinkedList<Integer> que = new LinkedList<>(); que.addLast(1); cost[1] = 0; vis[1] = true; while(!que.isEmpty()){ int cur = que.pollFirst(); for(int i=1;i<=cur;i++){ int nxt = cur + cur / i; if(nxt >= N){ continue; } if(vis[nxt]){ continue; } vis[nxt] = true; cost[nxt] = cost[cur] + 1; que.addLast(nxt); } } return cost; } public static void main(String[] args) { int[] cost = get_cost(); int[] b = new int[N]; int[] c = new int[N]; int[] w = new int[N]; int[] dp = new int[12 * N]; Scanner scanner = new Scanner(System.in); int T = scanner.nextInt(); while (T > 0) { int n = scanner.nextInt(); int k = scanner.nextInt(); for (int i = 1; i <= n; i++) { b[i] = scanner.nextInt(); } for (int i = 1; i <= n; i++) { c[i] = scanner.nextInt(); } for (int i = 1; i <= n; i++) { w[i] = cost[b[i]]; } int sum_w = 0; for (int i = 1; i <= n; i++) { sum_w += w[i]; } int sum_c = 0; for (int i = 1; i <= n; i++) { sum_c += c[i]; } if (k >= sum_w) { System.out.println(sum_c); } else { for (int i = 0; i <= k; i++) { dp[i] = 0; } for (int i = 1; i <= n; i++) { for (int j = k; j >= w[i]; j--) { int v = dp[j - w[i]] + c[i]; if (dp[j] < v) { dp[j] = v; } } } System.out.println(dp[k]); } T--; } } }

for i in range(1, cur+1): nxt = cur + cur // i if nxt >= N or vis[nxt]: continue vis[nxt] = True cost[nxt] = cost[cur] + 1 que.append(nxt) return cost def main(): cost = get_cost() dp = [0]...

#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include<string.h> #include<ctype.h> int main() { int n; int a[6002] = { 0 }; scanf("%d", &n); int sum = n; int flag = 0; int blag = 0; int count = 0; for (int i = 1; i <= n; i++) { a[i] = 1; } int j = 3, k = 2; while (sum > 3) { if (flag == 0) { for (int i = 1; i <= n; i++) { count++; if (count % 3 == 0) { if (a[i] == 2) { count--; } else { a[i] = 2; sum--; count = 0; } } } flag = 1; } if (flag == 1) { for (int i = 1; i <= n; i++) { count++; if (count % 2 == 0) { if (a[i] == 2) { count--; } else { a[i] = 2; sum--; count = 0; } } } flag = 0; } } for (int i = 1; i <= n; i++) { if (a[i] != 0 && a[i] != 2) { if (blag == 0) { printf("%d", i); blag = 1; } else printf(" %d", i); } } return 0; }这个代码怎么改

1. 增加注释来解释代码的功能和实现方式，提高代码可读性。 2. 将变量名改为有意义的名称，易于理解。 3. 将常量和变量的定义放在函数开头，方便查看。 4. 将一些重复的代码封装成函数，提高代码的复用性和可维护...

将下面这段代码翻译成python语言#include<stdio.h> #include<math.h> int pailie(int a,int b) { int ans=1; for(int i=a;i>(a-b);i--) { ans=i; } return ans; } int function(int m,int n){ if(m==n) { return 1; } int k,t,M=1,N=1,T=1; t=m-n; while(t!=1){ T=Tt; t=t-1; } while(m!=1){ M=Mm; m=m-1; } while(n!=1){ N=Nn; n=n-1; } k=M/(NT); return k; } int fact(int N) { int ans=1; if(N==0){ } else{ for(int i=1;i<=N;i++) { ans=i; } } return ans; } int DP(int a,int b) { int ans; return ans=fact(a)/fact(a-b); } int main() { int N,n; scanf("%d",&N); //a不唯一 for(int n=2;n<=N;n++){ int P[n]={0},F[n],sum1=0; double m[n]={0}; for(int i=0;i<n;i++) { F[i]=0; P[i]=DP(n,i+1); for(int k=0;k<=i+1;k++) { F[i]+=DP(i+1,k); } sum1+=P[i](F[i]-1); } for(int i=0;i<n;i++) { m[i]=(double)(F[i]-1)/(double)sum1; //printf("%lf\n",m[i]); } int cnt[n]={0},po[n]={0},sum=0; double fenzisum=0,fenmusum=0,tmp=0; for(int i=0;i<n;i++) { cnt[i]=pailie(n,i+1); //printf("%d %d %d\n",n,i+1,cnt[i]); } for(int i=0;i<n;i++) { po[i]=pow(2,(i+1))-1; sum+=cnt[i]po[i]; } for(int i=0;i<n;i++) { m[i]=(double)po[i]/sum; fenzisum+=m[i]log2(m[i]/po[i])cnt[i]; fenmusum+=pow(po[i],m[i])*cnt[i]; //printf("%lf\n",m[i]); } double ans=fenzisum/log2(fenmusum); printf("A=1 N=%d\n",n); printf("%lf\n",-ans);} }

P, F, sum1, m, cnt, po, sum, fenzisum, fenmusum = [0]*n, [0]*n, 0, [0]*n, [0]*n, [0]*n, 0, 0, 0 for i in range(n): P[i] = DP(n, i+1) for k in range(i+2): F[i] += DP(i+1, k) sum1 += P[i] * (F[i]-...

为代码写注释#include<stdio.h> #include<math.h> int main() { int x, N, c = 0; while (scanf("%d %d", &x, &N) == 2) { int arr[105]; int flag = 0; for (int i = 1; i <= N; i++) arr[i] = pow(i, x); printf("Case %d:\n", ++c); for (int i = 1; i <= N; i++) { for (int j = i; j <= N; j++) { for (int k = j; k <= N; k++) { int sum = arr[i] + arr[j] + arr[k]; for (int t = k + 1; t <= N; t++) { if (sum == arr[t]) { flag = 1; printf("%d^%d+%d^%d+%d^%d=%d^%d\n", i, x, j, x, k, x, t, x); break; } if (arr[t] > sum)break; } } } } if (!flag) printf("No such numbers.\n"); } return 0; }

i <= N; i++) // 循环计算 i^x 的值 arr[i] = pow(i, x); // 使用 pow() 函数计算 i 的 x 次方，存储在数组 arr 中 printf("Case %d:\n", ++c); // 输出当前测试用例的编号 for (int i = 1; i <= N; i++) { // ...

#include <bits/stdc++.h> using namespace std; void solve(int w[],int n,int k){ int sum=0; int count=0; for(int i=0;i<=k;i++){ if(w[i]>n){ cout<<"No Solution"<<endl; return; } } for(int i=0;i<=k;i++){ sum+=w[i]; if(sum>n){ count++; sum=w[i]; } } cout<<count<<endl; } int main(){ int n=0; int k=0; while(cin>>n){ cin >>k; int way[k+1]; for(int i=0;i<k+1;i++){ cin>> way[i]; } solve(way,n,k); } }转java

i <= k; i++) { w[i] = scanner.nextInt(); } solve(w, n, k); } scanner.close(); } private static void solve(int[] w, int n, int k) { int sum = 0; int count = 0; for (int i = 0; i <= k; i++) ...

include <iostream.h> void main( ) { int i=1,sum=0; while(i<10) i++; sum=sum+i; cout<<i<<" "<<sum<<endl; }

这段代码中存在一个错误，... int i=1, sum=0; while(i<10) { sum=sum+i; i++; } cout<<i<<" "<<sum<<endl; } 这段代码的功能是计算 1 到 9 的和，并输出结果。修改后的代码可以正确地实现这个功能。

解释以下代码：#include <bits/stdc++.h> using namespace std; struct Node { int id,f; bool operator < (const Node &a)const{ return f<a.f; } } a[105]; int n,h,ans; int d[105],t[105]; priority_queue <Node> q; int main() { cin>>n; for(int i=1; i<=n; i++) { cin>>a[i].f; a[i].id=i; } for(int i=1; i<=n; i++)cin>>d[i]; t[1]=0; for(int i=2; i<=n; i++)cin>>t[i]; cin>>h; for(int i=1; i<=n; i++) { h-=t[i]; for(int j=1; j<=i; j++) q.push(a[j]); int m=h,sum=0; while(m>0) { Node a=q.top(); q.pop(); if(a.f<=0) break; sum+=a.f; a.f-=d[a.id]; q.push(a); m--; } ans=max(ans,sum); } cout<<ans; return 0; 你对于信息奥赛一本通的理解

否则，将该数据的得分加到sum中，减去对应的消耗d[a.id]，然后将更新后的数据重新加入优先队列q中，并将剩余可用时间m减1。最后，将当前得分ans与sum进行比较，取较大值。最后输出最大得分ans。这段代码的目的是...

int main() { int n,m,i,sum=0; cin>>n>>m; for( i=49;i<=m;i++) { sum+=i%10; i=i/10; sum=sum+i; if(sum==13) cout<<i<<endl; } return 0; }

Note that there is a syntax error in the program: the loop condition should be i<=m instead of i to avoid an infinite loop. Additionally, there is no need to declare the variable sum outside of the ...

将以下java代码转换为C++代码：public class { public static void main(String[] args) { int j=0; for (int i=1;i<10000000;i++){ if (BaseConversion(i)){ j++; if (j==2023){ System.out.println(i);//215040 break; } } } } public static boolean BaseConversion(int n){ //十进制 int sum=0; int x=n; while (x!=0){ sum+=(x%10); x/=10; } if (n%sum!=0) return false; //二进制 sum=0; x=n; while (x!=0){ sum+=(x%2); x/=2; } if (n%sum!=0) return false; //八进制 sum=0; x=n; while (x!=0){ sum+=(x%8); x/=8; } if (n%sum!=0) return false; //十六进制 int[] arr={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}; sum=0; x=n; while (x!=0){ sum+=(arr[x%16]); x/=16; } if (n%sum!=0) return false; return true; } }

int sum = 0; int x = n; while (x != 0) { sum += (x % 10); x /= 10; } if (n % sum != 0) return false; // 二进制 sum = 0; x = n; while (x != 0) { sum += (x % 2); x /= 2; } if (n % sum !...

#include<iostream> using namespace std; const int N=1e5+10; long long a[N],n,m,sum,maxa; int check(int mid) { int sum=0; for(int i=1;i<=n;i++){ sum+=a[i]/mid; } if(sum>=m) return 1; //总段数大于等于所需要的 return 0; } int main(){ cin>>n>>m; for(int i=1;i<=n;i++){ cin>>a[i],sum+=a[i]; if(a[i]>maxa) maxa=a[i]; } if(sum<m){cout<<0;return 0;} //先判断是否有解 int l=1,r=maxa; while(l<r) //模板2 { int mid=l+r+1>>1; if(check(mid)) l=mid; else r=mid-1; } cout<<l; return 0; }

如果 sum 小于 m，表示总和不足以满足所需总段数，输出 0 并结束程序。若存在解决方案，程序使用二分查找的模板2来确定一个最小的满足条件的值。循环中，定义了一个中间值 mid，并通过调用 check 函数来判断 mid ...

#include<algorithm> #include<iostream> using namespace std; const int N = 1e6; int arr[N]; int atr[N]; int k, c, n; int sum = 0; int main() { int t; cin >> t; while (t--) { cin >> k >> c >> n; for (int i = 1; i <= n; i++) { cin >> arr[i]; sum += arr[i]; } sort(arr + 1, arr + n + 1); int gg = c; while (k) { if (n == 0)break; int l = 1, r = n; while (l < r) { int mid = (l + r + 1) >> 1; if (arr[mid] < gg) { l = mid; } else { r = mid - 1; } } if (arr[l] <= gg) { gg -= arr[l]; int b[200]; int cnt = 1; // arr[l] = 0; for (int i = 1; i <= n; i++) if (i != l)b[cnt++] = arr[i]; n--; for (int i = 1; i <= n; i++)arr[i] = b[i]; sort(arr + 1, arr + 1 + n); } else if (k != 1) { gg = c; k--; } else { k--; } } if (n == 0)cout << "Let's go\n"; else cout << "find another way\n"; } return 0; }优化上述代码

for (int i = 0; i < n; ++i) { cin >> arr[i]; sum += arr[i]; } sort(arr, arr + n); int gg = c; bool success = true; for (int i = 0; i < k; ++i) { auto iter = lower_bound(arr, arr + n, gg);...

补全以下这段代码 using namespace std; const int N = 51; int n,m,a[N],k,x; unordered_map<int,int> mp; void dfs(int x,int num,int sum) { if (num==4) { mp[sum]=1; return; } for (int i=x+1;i<=n+num-3;i++) dfs(i,num+1,sum+a[i]); } inline void Main() { cin>>n>>m; for (int i=1;i<=n;i++) cin>>a[i]; dfs(0,0,0); while (m--) { cin>>k; bool boo=true; for (int i=1;i<=k;i++) { cin>>x; if (mp[x*4]) continue; boo=false; } if (boo) cout<<"Yes"<<endl; else cout<<"No"<<endl; } }

i <= n + num - 3; i++) dfs(i, num + 1, sum + a[i]); } inline void Main() { cin >> n >> m; for (int i = 1; i <= n; i++) cin >> a[i]; dfs(0, 0, 0); while (m--) { cin >> k; bool boo = true; ...

优化程序：#include<bits/stdc++.h> using namespace std; int a[200005],b[200005],c[200005]; int main() { int n,m,sum=1e9; cin>>n>>m; for(int i=1;i<=n;i++){ cin>>a[i]; } for(int i=1;i<=m;i++){ cin>>b[i]; } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(a[i]>b[j]) sum = min(sum,a[i]-b[j]); if(a[i]<b[j]) sum = min(sum,b[j]-a[i]); if(a[i]==b[j]){ cout<<0; return 0; } } } cout<<sum; return 0; }

while(i < n && j < m){ minDiff = min(minDiff, abs(a[i] - b[j])); if(a[i] < b[j]){ i++; } else{ j++; } } cout << minDiff << endl; return 0; } 这个优化后的代码将遍历一次数组，找到...

int main() { int x=1; int i =0; int n=0; int sum=0; while(n<=10) { for(i =1;i<=n;i++) { x*=i; sum=sum+x; } n++; } Return 0; }

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C++实现1-n连续整数求和的简易代码解析

C语言实现1至100累加之和的while循环算法

main()int i=1,sum=0while (i <= 100) sum =sum+1; i++;while (i <= 10) sum =sum+1; i++;print("i=％d，sum=％d",i,sum)

include <iostream.h> void main( ) { int i=1,sum=0; while(i<10) i++; sum=sum+i; cout<<i<<" "<<sum<<endl; }

int main() { int n,m,i,sum=0; cin>>n>>m; for( i=49;i<=m;i++) { sum+=i%10; i=i/10; sum=sum+i; if(sum==13) cout<<i<<endl; } return 0; }

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