[NOIP2013] 普及组

计数问题

纯模拟

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int main(){
 5     int n,x;
 6     cin>>n>>x;
 7     int i,c=0;
 8     for(i=1;i<=n;i++){
 9         int a=i;
10         while(a!=0){
11             if(a%10==x)c++;
12             a/=10;
13         }
14     }
15     cout<<c;
16     return 0;
17 }

计数问题

表达式求值

模拟计算，扫一遍出解

 1 /*NOIP2013普及组t2 洛谷P1981 表达式求值*/
 2 /**/
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<cstdio>
 7 #include<cmath>
 8 using namespace std;
 9 char last;
10 char c;
11 int x=0;
12 int a=0,b=1;
13 int sum=0;
14 int main(){
15     int i,j;
16     bool flag=1;
17     do{
18         if(cin>>c);
19         else{
20             flag=0;
21             c=‘+‘;
22         }
23         if(c>=‘0‘ && c<=‘9‘)x=x*10+c-‘0‘;
24         else{
25             a=x;
26             x=0;
27         }
28         if(c==‘*‘){
29             last=1;
30             b=(a*b)%10000;
31         }
32         if(c==‘+‘){
33             if(last){
34                 a=(a*b)%10000;
35                 sum=(sum+a)%10000;
36                 b=1;
37                 last=0;
38             }
39             else sum+=a;
40         }
41
42     }while(flag==1);
43     printf("%d",sum%10000);
44     return 0;
45 }

表达式求值

小朋友的数字

线性DP

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 using namespace std;
 6 const long long inf=1000000005;
 7 int n,p,a[1000500];
 8
 9 long long su[1000500]={0};//特征值
10 long long scoremx=-inf;//前排最大分数
11 long long dmx=-inf;
12 long long ans=-inf;
13 int sum1(){
14     int i,j;
15     long long s=0;
16     long long mx=-inf;
17     for(i=1;i<=n;i++){
18         if(s+a[i]>mx)mx=s+a[i];
19         su[i]=mx;
20         if(s+a[i]>0) s+=a[i];
21         else s=0;
22     }
23     //
24     for(i=1;i<=n;i++){
25         su[i]%=p<<1;
26     }
27     return 0;
28 }
29 int main(){
30     int i,j;
31     scanf("%d%d",&n,&p);
32     for(i=1;i<=n;i++)scanf("%d",&a[i]);
33     sum1();
34     long long score;
35 //1
36     scoremx=su[1];
37     score=su[1];
38     dmx=score+su[1];
39 //end
40     for(i=2;i<=n;i++){
41         score=dmx;
42         if(dmx>scoremx)scoremx=dmx;
43         if(su[i]+score>dmx)dmx=su[i]+score;
44     }
45     printf("%d",scoremx%p);
46     return 0;
47 }

小朋友的数字

车站分级

从已知的低级站向高级站连边，跑拓扑排序得出层级关系

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 using namespace std;
 7 int n,m;
 8 int a;//[第i趟车次的停靠站数]
 9 int s[2000];//[第i趟车次停靠的站]
10 int mp[1200][1200]={0};
11 int book[1200];
12 int r[1200],c[1200];
13 int st[1200];
14 int ans=0;
15 void rd(){
16     scanf("%d%d",&n,&m);
17     int i,j,k;
18     for(i=1;i<=m;i++){
19         memset(book,0,sizeof(book));
20         scanf("%d",&a);
21         for(j=1;j<=a;j++){
22             scanf("%d",&s[j]);
23             book[s[j]]=1;
24         }
25         for(k=s[1];k<=s[a];k++)//遍历从始发站到终点站
26         {
27             if(!book[k])
28                 for(j=1;j<=a;j++)
29                 if(!mp[k][s[j]])//从低级连到高级可过，从高级连到低级超时
30                 {
31                  mp[k][s[j]]=1;
32                  r[s[j]]++;}
33         }
34
35     }
36 }
37 int main(){
38 //    freopen("level.in","r",stdin);
39 //    freopen("level.out","w",stdout);
40     rd();
41     int i,j;
42     int top=0;
43     memset(book,0,sizeof(book));
44     while(1){
45         ans++;
46         top=0;
47         for(i=1;i<=n;i++)
48             if(!r[i] && !book[i])
49             {
50                 top++;
51                 st[top]=i;
52                 book[i]=1;
53             }
54
55         if(!top)break;
56         for(j=1;j<=n;j++)
57           for(i=1;i<=top;i++){
58               if(mp[st[i]][j]){
59                   mp[st[i]][j]=0;
60                   r[j]--;
61               }
62           }
63         if(!top)break;
64     }
65
66     printf("%d",ans-1);
67
68
69     return 0;
70 }