https://codeforces.com/contest/1213
给定n个数，每个数加减2不需要花费，加减1需要花费1个coin，求将这n个数变为同一个数的最小花费
只需要求偶数个数和奇数个数即可，取较小值
1 #include<iostream> 2 #include<sstream> 3 #include<fstream> 4 #include<algorithm> 5 #include<cstring> 6 #include<iomanip> 7 #include<cstdlib> 8 #include<cctype> 9 #include<vector> 10 #include<string> 11 #include<cmath> 12 #include<ctime> 13 #include<stack> 14 #include<queue> 15 #include<map> 16 #include<set> 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define random(a,b) (rand()%(b-a+1)+a) 19 #define ll long long 20 #define ull unsigned long long 21 #define e 2.71828182 22 #define Pi acos(-1.0) 23 #define ls(rt) (rt<<1) 24 #define rs(rt) (rt<<1|1) 25 #define lowbit(x) (x&(-x)) 26 using namespace std; 27 int n,odd,even,t; 28 int read() 29 { 30 int s=1,x=0; 31 char ch=getchar(); 32 while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();} 33 while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();} 34 return x*s; 35 } 36 int main() 37 { 38 n=read(); 39 for(int i=1;i<=n;++i) 40 { 41 t=read(); 42 if(t&1) odd++; 43 else even++; 44 } 45 cout<<min(odd,even)<<endl; 46 }
给定一个序列，定义一个元素为bad为存在比它小的元素在它后面，求bad元素的个数
逆序，树状数组求
记得离散化
1 #include<iostream> 2 #include<sstream> 3 #include<fstream> 4 #include<algorithm> 5 #include<cstring> 6 #include<iomanip> 7 #include<cstdlib> 8 #include<cctype> 9 #include<vector> 10 #include<string> 11 #include<cmath> 12 #include<ctime> 13 #include<stack> 14 #include<queue> 15 #include<map> 16 #include<set> 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define random(a,b) (rand()%(b-a+1)+a) 19 #define ll long long 20 #define ull unsigned long long 21 #define e 2.71828182 22 #define Pi acos(-1.0) 23 #define ls(rt) (rt<<1) 24 #define rs(rt) (rt<<1|1) 25 #define lowbit(x) (x&(-x)) 26 using namespace std; 27 const int MAXN=150000+5; 28 int T[MAXN],a[MAXN],b[MAXN]; 29 int n,t; 30 int read() 31 { 32 int s=1,x=0; 33 char ch=getchar(); 34 while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();} 35 while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();} 36 return x*s; 37 } 38 int getsum(int x) 39 { 40 int ans=0; 41 for(int i=x;i>=1;i-=lowbit(i)) 42 ans+=T[i]; 43 return ans; 44 } 45 void update(int x,int v) 46 { 47 for(int i=x;i<=n;i+=lowbit(i)) 48 T[i]+=v; 49 } 50 int main() 51 { 52 t=read(); 53 while(t--) 54 { 55 mem(T,0); 56 n=read(); 57 for(int i=1;i<=n;++i) 58 a[i]=read(),b[i]=a[i]; 59 sort(b+1,b+n+1); 60 int res=0; 61 for(int i=n;i>=1;--i) 62 { 63 a[i]=lower_bound(b+1,b+n+1,a[i])-b; 64 if(getsum(a[i]-1)) res++; 65 update(a[i],1); 66 } 67 cout<<res<<endl; 68 } 69 }
emmm，其实想复杂了，逆序求，维护最小值就行，O(n)复杂度
1 #include<iostream> 2 #include<sstream> 3 #include<fstream> 4 #include<algorithm> 5 #include<cstring> 6 #include<iomanip> 7 #include<cstdlib> 8 #include<cctype> 9 #include<vector> 10 #include<string> 11 #include<cmath> 12 #include<ctime> 13 #include<stack> 14 #include<queue> 15 #include<map> 16 #include<set> 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define random(a,b) (rand()%(b-a+1)+a) 19 #define ll long long 20 #define ull unsigned long long 21 #define e 2.71828182 22 #define Pi acos(-1.0) 23 #define ls(rt) (rt<<1) 24 #define rs(rt) (rt<<1|1) 25 #define lowbit(x) (x&(-x)) 26 using namespace std; 27 const int MAXN=150000+5; 28 int a[MAXN]; 29 int n,t; 30 int read() 31 { 32 int s=1,x=0; 33 char ch=getchar(); 34 while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();} 35 while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();} 36 return x*s; 37 } 38 int main() 39 { 40 t=read(); 41 while(t--) 42 { 43 n=read(); 44 for(int i=1;i<=n;++i) 45 a[i]=read(); 46 int res=0,small=1<<30; 47 for(int i=n;i>=1;--i) 48 { 49 if(a[i]>small) res++; 50 small=min(small,a[i]); 51 } 52 cout<<res<<endl; 53 } 54 }
对于每个n和m，求[1,n]中除以0余数为0的数的末尾数的和。
找规律
1 #include<iostream> 2 #include<sstream> 3 #include<fstream> 4 #include<algorithm> 5 #include<cstring> 6 #include<iomanip> 7 #include<cstdlib> 8 #include<cctype> 9 #include<vector> 10 #include<string> 11 #include<cmath> 12 #include<ctime> 13 #include<stack> 14 #include<queue> 15 #include<map> 16 #include<set> 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define random(a,b) (rand()%(b-a+1)+a) 19 #define ll long long 20 #define ull unsigned long long 21 #define e 2.71828182 22 #define Pi acos(-1.0) 23 #define ls(rt) (rt<<1) 24 #define rs(rt) (rt<<1|1) 25 #define lowbit(x) (x&(-x)) 26 using namespace std; 27 ll read() 28 { 29 ll s=1,x=0; 30 char ch=getchar(); 31 while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();} 32 while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();} 33 return x*s; 34 } 35 int main() 36 { 37 ll q=read(); 38 ll T,ans; 39 while(q--) 40 { 41 ll n=read(),m=read(); 42 if(m%10==5) 43 { 44 T=n/(2*m); 45 ans=T*5; 46 for(ll i=2*T*m;i<=n;i+=m) 47 if(i%m==0) ans+=i%10; 48 } 49 else if(m%10==0) ans=0; 50 else if(m%2==1) 51 { 52 T=n/(m*10); 53 ans=T*45; 54 for(ll i=10*T*m;i<=n;i+=m) 55 if(i%m==0) ans+=i%10; 56 } 57 else if(m%2==0) 58 { 59 T=n/(m*5); 60 ans=T*20; 61 for(ll i=5*T*m;i<=n;i+=m) 62 if(i%m==0) ans+=i%10; 63 } 64 cout<<ans<<endl; 65 } 66 }