标签：bsp art mat eve algorithm lock lag 技术 表达式求值
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 }
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 }
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 }
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 }
标签：bsp art mat eve algorithm lock lag 技术 表达式求值
原文地址：http://www.cnblogs.com/SilverNebula/p/6067469.html
