A - Second Half Sum题目描述给你一个长度为 $N$ 的整数序列 $A(A_1,A_2,\dots,A_N)$ .这里 $N$ 是偶数。 求 $A$ 的后半部之和即 $A_{(N/2)1},A_{(N/2)2},\dots,A_N$ 的和。解题思路A题不讲了。脑残题。完整代码#includebits/stdc.h #define fr1(i,a,b) for(int (i)(a);(i)(b);(i)) #define fr2(i,a,b) for(int (i)(a);(i)(b);--(i)) #define fr3(i,a,b,n) for(int (i)(a);(i)(b);(i)(n)) #define fr4(i,a,b,n) for(int (i)(a);(i)(b);(i)-(n)) #define fv(i,p) for(auto (i):(p)) #define ll long long #define ull unsigned ll #define pii pairint,int #define pll pairll,ll #define _1st first #define _2nd second #define y1 yy1 #define elif else if #define RT return #define debug coutendl-------------------------------------------------------------endl using namespace std; int n,ans; int main(){ ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL); cinn; fr1(i,1,n){ int x; cinx; if(in/2)ansx; } coutans; RT 0; }B - Old Maid高桥目前有 $N$ 张牌。其中 第$i$ $(1\le i\le N)$ 张牌上写着整数 $A_i$ 。他尽可能多地重复下面的操作。选择两张写有相同整数的不同卡片然后吃掉这两张卡片。被吃掉的牌将被永久删除无法在后续操作中选择。求当无法再进行操作时写在卡片上的整数之和。解题思路卡片能吃排序。完整代码#includebits/stdc.h #define fr1(i,a,b) for(int (i)(a);(i)(b);(i)) #define fr2(i,a,b) for(int (i)(a);(i)(b);--(i)) #define fr3(i,a,b,n) for(int (i)(a);(i)(b);(i)(n)) #define fr4(i,a,b,n) for(int (i)(a);(i)(b);(i)-(n)) #define fv(i,p) for(auto (i):(p)) #define ll long long #define ull unsigned ll #define pii pairint,int #define pll pairll,ll #define _1st first #define _2nd second #define y1 yy1 #define elif else if #define RT return #define debug coutendl-------------------------------------------------------------endl using namespace std; int n,a[114514],ans; int main(){ ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL); cinn; fr1(i,1,n){ cina[i]; } sort(a1,an1); fr1(i,1,n){ if(a[i]a[i1]){ a[i]0;a[i1]0; } }fr1(i,1,n){ ansa[i]; }coutans; RT 0; }C - Change Schools题目描述目前AtCoder 高中有 $K$ 个班级和 $N$ 名学生其中 $i$ 个学生属于 $(1\le i\le N)$ 个班级。高桥将在九月份转入 AtCoder 高中。届时他可以在 $K$ 个班级中任意选择一个班级并属于那个班级。如果有一个班级的学生比他所在的班级多他就会伤心。否则他会很高兴。求如果他属于多少个班他就会感到快乐。解题思路最优策略把新增学生放到一个人数 mx‑1 的班级这样得到最多的等于最大值的班级。所以答案统计有多少个班级满足 \(cnt_i1 \ge mx\)。\(cnt_i mx\)\(cnt_i1 \ge mx\) 成立这些班本身就是最大值\(cnt_i mx‑1\)\(cnt_i1 mx\)加一个人之后也成为最大值\(cnt_i \le mx‑2\)\(cnt_i1 \le mx‑1\)达不到最大值不计入。完整代码#includebits/stdc.h #define fr1(i,a,b) for(int (i)(a);(i)(b);(i)) #define fr2(i,a,b) for(int (i)(a);(i)(b);--(i)) #define fr3(i,a,b,n) for(int (i)(a);(i)(b);(i)(n)) #define fr4(i,a,b,n) for(int (i)(a);(i)(b);(i)-(n)) #define fv(i,p) for(auto (i):(p)) #define ll long long #define ull unsigned ll #define pii pairint,int #define pll pairll,ll #define _1st first #define _2nd second #define y1 yy1 #define elif else if #define RT return #define debug coutendl-------------------------------------------------------------endl using namespace std; int n,k,a[200001],ans; pii cl[200001];//班级人数 int main(){ ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL); cinnk; fr1(i,1,k){ cl[i]._1sti; cl[i]._2nd0; } fr1(i,1,n){ cina[i]; cl[a[i]]._2nd; } int mx0; fr1(i,1,k){ mxmax(mx,cl[i]._2nd); } fr1(i,1,k){ if(cl[i]._2nd1mx)ans; } coutans; RT 0; }D - Coefficient Stair题目描述输出由满足 $\displaystyle\sum_{i1} ^ Ni\times A_iK$ 的非负整数组成的所有长度为 $N$ 的序列 $A(A_1,A_2,\ldots,A_N)$ 按字典序排列。解题思路使用暴力打表的方式枚举前n-1个数最后一个变量直接通过等式算出来\(n\cdot a_n k - \sum_{i1}^{n‑1}i\cdot a_i\)完整代码#includebits/stdc.h #define fr1(i,a,b) for(int (i)(a);(i)(b);(i)) #define fr2(i,a,b) for(int (i)(a);(i)(b);--(i)) #define fr3(i,a,b,n) for(int (i)(a);(i)(b);(i)(n)) #define fr4(i,a,b,n) for(int (i)(a);(i)(b);(i)-(n)) #define fv(i,p) for(auto (i):(p)) #define ll long long #define ull unsigned ll #define pii pairint,int #define pll pairll,ll #define _1st first #define _2nd second #define y1 yy1 #define elif else if #define RT return #define debug coutendl-------------------------------------------------------------endl using namespace std; int n,k; int main(){ ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL); cinnk; if(n1){ cout k \n; }elif(n2){ fr1(a1,0,k){ if((k-1*a1)%20){ couta1 (k-1*a1)/2\n; } } }elif(n3){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ if((k-1*a1-2*a2)%30){ couta1 a2 (k-1*a1-2*a2)/3\n; } } } }elif(n4){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ if((k-1*a1-2*a2-3*a3)%40){ couta1 a2 a3 (k-1*a1-2*a2-3*a3)/4\n; } } } } }elif(n5){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ fr1(a4,0,(k-1*a1-2*a2-3*a3)/4){ if((k-1*a1-2*a2-3*a3-4*a4)%50){ couta1 a2 a3 a4 (k-1*a1-2*a2-3*a3-4*a4)/5\n; } } } } } }elif(n6){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ fr1(a4,0,(k-1*a1-2*a2-3*a3)/4){ fr1(a5,0,(k-1*a1-2*a2-3*a3-4*a4)/5){ if((k-1*a1-2*a2-3*a3-4*a4-5*a5)%60){ couta1 a2 a3 a4 a5 (k-1*a1-2*a2-3*a3-4*a4-5*a5)/6\n; } } } } } } }elif(n7){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ fr1(a4,0,(k-1*a1-2*a2-3*a3)/4){ fr1(a5,0,(k-1*a1-2*a2-3*a3-4*a4)/5){ fr1(a6,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5)/6){ if((k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6)%70){ couta1 a2 a3 a4 a5 a6 (k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6)/7\n; } } } } } } } }elif(n8){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ fr1(a4,0,(k-1*a1-2*a2-3*a3)/4){ fr1(a5,0,(k-1*a1-2*a2-3*a3-4*a4)/5){ fr1(a6,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5)/6){ fr1(a7,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6)/7){ if((k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7)%80){ couta1 a2 a3 a4 a5 a6 a7 (k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7)/8\n; } } } } } } } } }elif(n9){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ fr1(a4,0,(k-1*a1-2*a2-3*a3)/4){ fr1(a5,0,(k-1*a1-2*a2-3*a3-4*a4)/5){ fr1(a6,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5)/6){ fr1(a7,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6)/7){ fr1(a8,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7)/8){ if((k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7-8*a8)%90){ couta1 a2 a3 a4 a5 a6 a7 a8 (k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7-8*a8)/9\n; } } } } } } } } } }elif(n10){ fr1(a1,0,k){ fr1(a2,0,(k-1*a1)/2){ fr1(a3,0,(k-1*a1-2*a2)/3){ fr1(a4,0,(k-1*a1-2*a2-3*a3)/4){ fr1(a5,0,(k-1*a1-2*a2-3*a3-4*a4)/5){ fr1(a6,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5)/6){ fr1(a7,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6)/7){ fr1(a8,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7)/8){ fr1(a9,0,(k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7-8*a8)/9){ if((k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7-8*a8-9*a9)%100){ couta1 a2 a3 a4 a5 a6 a7 a8 a9 (k-1*a1-2*a2-3*a3-4*a4-5*a5-6*a6-7*a7-8*a8-9*a9)/10\n; } } } } } } } } } } } RT 0; }