寒假第二周 1.18 --- 1.24

1.18

cf 581c

581C - Developing Skills

重新自己写了一遍，注意都是0 的时候

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<vector>
 6 using namespace std;
 7
 8 const int maxn = 1e5+5;
 9 int n,k;
10
11 struct node{
12     int x,y;
13 }p[maxn];
14
15 int cmp(node n1,node n2){
16     return n1.y < n2.y;
17 }
18
19 int cmp0(node n1,node n2){
20     return n1.x < n2.x;
21 }
22
23 int a[105];
24
25 void solve(){
26     for(int i = 1;i <= n;i++){
27         int pos = lower_bound(a+1,a+10,p[i].x) - a;
28         if(a[pos] == p[i].x) pos++;
29         p[i].y = a[pos] - p[i].x;
30      //  printf("a[%d] = %d  p[%d].x = %d\n",pos,a[pos],i,p[i].x);
31     }
32    /* for(int i = 1;i <= n;i++){
33         printf("p[%d].x = %d  y = %d\n",i,p[i].x,p[i].y);
34     }*/
35     sort(p+1,p+n+1,cmp);
36     int ans = 0;
37     for(int i = 1;i <= n;i++){
38         if(k >= p[i].y){
39             ans += (p[i].x + p[i].y)/10;
40             k -= p[i].y;
41             p[i].x = p[i].x + p[i].y;
42         }
43         else {
44             ans += p[i].x/10;
45         }
46        // printf("i = %d  k = %d  ans = %d\n",i,k,ans);
47     }
48     if(k){
49         for(int i = 1;i <= n;i++){
50             int l = 10 - p[i].x/10;
51             int r = k/10;
52             if(r >= l){
53                 ans += l;
54                 k = k-l*10;
55             }
56             else{
57                 ans += k/10;
58                 k = k%10;
59             }
60             if(k < 10) break;
61             //printf("l = %d  r = %d  k = %\n",l,r,k);
62         }
63     }
64     printf("%d\n",ans);
65 }
66
67 int main(){
68     for(int i = 1;i <= 10;i++) a[i] = i*10;
69     a[11] = 100;
70   //  freopen("in.txt","r",stdin);
71    // freopen("out.txt","w",stdout);
72     while(scanf("%d %d",&n,&k) != EOF){
73         for(int i = 1;i <= n;i++){
74             scanf("%d",&p[i].x);
75         }
76         solve();
77     }
78     return 0;
79 }

cf 614e

614E - Necklace

先不理解题解说的 part 是什么意思，，就是所有a[i] 的gcd

然后像题解说的那样构造

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<vector>
  4 #include<iostream>
  5 #include<algorithm>
  6 using namespace std;
  7
  8 const int maxn = 1e5+5;
  9 char s[maxn];
 10 int n,a[maxn],aa[maxn];
 11
 12 int gcd(int a,int b){
 13     return (!b) ? a:gcd(b,a%b);
 14 }
 15
 16 void solve(){
 17     int part = a[1];
 18     for(int i = 2;i <= n;i++){
 19         part = gcd(part,a[i]);
 20     }
 21     for(int i = 1;i <= n;i++){
 22         aa[i] = a[i]/part;
 23     }
 24
 25     if(part%2 == 0){
 26         printf("%d\n",part);
 27         vector<char> c;
 28         for(int i = 1;i <= n;i++){
 29             for(int j = 1;j <= aa[i];j++){
 30                 c.push_back(i+‘a‘-1);
 31             }
 32         }
 33         int flag = 1;
 34         for(int i = 1;i <= part;i++){
 35             if(flag){
 36                 for(int j = 0;j < c.size();j++){
 37                     printf("%c",c[j]);
 38                 }
 39             }
 40             else{
 41                 for(int j = c.size()-1;j >= 0;j--){
 42                     printf("%c",c[j]);
 43                 }
 44             }
 45             flag = !flag;
 46         }
 47         printf("\n");
 48         return;
 49     }
 50     int ji = 0;
 51     for(int i = 1;i <= n;i++){
 52         ji += (aa[i]%2);
 53     }
 54     if(ji > 1){
 55         puts("0");
 56         for(int i = 1;i <= n;i++){
 57             for(int j = 1;j <= a[i];j++){
 58                 char z = i+‘a‘-1;
 59                 printf("%c",z);
 60             }
 61         }
 62         printf("\n");
 63     }
 64     if(ji == 1){
 65         vector<char> l;
 66         vector<char> r;
 67         int pos;
 68         for(int i = 1;i <= n;i++){
 69             if(a[i]%2){
 70                 pos = i;
 71                 break;
 72             }
 73         }
 74         for(int i = 1;i <= n;i++){
 75             if(i == pos) continue;
 76             for(int j = 1;j <= aa[i]/2;j++){
 77                 l.push_back(i+‘a‘-1);
 78                 r.push_back(i+‘a‘-1);
 79             }
 80         }
 81         for(int j = 1;j <= aa[pos];j++){
 82             l.push_back(pos+‘a‘-1);
 83         }
 84         reverse(r.begin(),r.end());
 85         for(int j = 0;j < r.size();j++){
 86             l.push_back(r[j]);
 87         }
 88         printf("%d\n",part);
 89         for(int i = 1;i <= part;i++){
 90             for(int j = 0;j < l.size();j++){
 91                 printf("%c",l[j]);
 92             }
 93         }
 94         printf("\n");
 95     }
 96 }
 97
 98 int main(){
 99     while(scanf("%d",&n) != EOF){
100         for(int i = 1;i <= n;i++){
101             scanf("%d",&a[i]);
102         }
103         solve();
104     }
105     return 0;
106 }