A

解法：DP+二分

dp[i]=max(dp[i],dp[j]+p[i].v)(i>j)

dp[i]表示建立i点之后能够获得的最大值

int n,M;

struct node

{

int l,v;

}p[1010];

int dp[1010];

bool cmp(node a,node b){

return a.l < b.l;

}

bool judge_oo(){

int Max = -1;

for(int i = 1;i <= n;i++) Max = max(Max,p[i].v);

if(Max >= M) return 1;

return 0;

}

bool judge_no(){

int sum = 0;

for(int i = 1;i <= n;i++) sum += p[i].v;

if(sum >= M) return 0;

return 1;

}

bool judge_DP(int L){

memset(dp,0,sizeof dp);

for(int i = 1;i <= n;i++){

for(int j = 0;j < i; j++){

if(p[i].l - p[j].l >= L){

dp[i] = max(dp[i],dp[j] + p[i].v);

}

}

}

int ML = 0;

for(int i = 1;i <= n;i++) ML = max(ML,dp[i]);

if(ML >= M) return 1;

return 0;

}

int solve(int L,int R){

while(L + 1 < R){

int mid = (L + R)/2;

if(judge_DP(mid)) L = mid;

else R = mid;

}

return L;

}

int main(int argc, char *argv[])

{

//freopen("data.in","r",stdin);

//freopen("data.out","w",stdout);

int T;

cin >> T;

while(T--){

cin >> n >> M;

for(int i = 1;i <= n;i ++)

cin >> p[i].l >> p[i].v;

if(judge_no()){

puts("-1");

continue;

}

if(judge_oo()){

puts("oo");

continue;

}

sort(p+1,p+n+1,cmp);

p[0].l = -100000000 , p[0].v = 0;

int L = 1 , R = p[n].l;

cout << solve(L,R) << endl;

}

return 0;

}

B

解法：模拟。字符串模拟

#include

//#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#define inf 0x3fffffff

#define INF 0x3f3f3f3f

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define ULL unsigned long long

using namespace std;

int i,j;

int n,m;

int num_1,num_2;

int t;

int sum,ans,flag;

string s_1,s_2,s_3;

string sss;

string ss[10000];

string c;

mapq1;

mapq2;

int main()

{

cin>>t;

while(t--)

{

cin>>n>>c;

num_1=0;

for(i=0; i

{

cin>>sss;

ans=sss.find(".");

s_1=sss.substr(0,ans);

s_2=sss.substr(ans+1,sss.length());

if(s_2==c)

{

int ans_2=sss.find("(");

int ans_3=sss.find(")");

if((ans_2!=-1&&ans_2

{

// num_1++;

ss[num_1++]=sss.substr(0,ans_2);

// cout<

}

else if(ans_2==-1||ans_3==-1)

{

// num_1++;

ss[num_1++]=sss.substr(0,ans);

// cout<

}

}

}

for(i=0; i

{

// cout<

}

// cout<

for(i=0; i

{

q1[ss[i]]++;

}

for(i=0; i

{

if(q2[ss[i]]==0)

{

q2[ss[i]]=1;

if(q1[ss[i]]==1)

{

cout<

}

else

{

cout<

}

}

}

q1.clear();

q2.clear();

}

return 0;

}

C

解法：树状数组寻找逆序对+预处理

#include

#include

#include

#include

#include

using namespace std;

const int M = 200010;

long long ans[M + 10];

int G[M + 10],c[M + 10];

int Lowbit(int x)

{

return x & (-x);

}

void Insert(int x,int d)

{

while(x<=M){

c[x]+=d;

x+=Lowbit(x);

}

}

int Sum(int x)

{

int sum=0;

while(x>0){

sum+=c[x];

x-=Lowbit(x);

}

return sum;

}

void init()

{

memset(G,-1,sizeof G);

memset(ans,0,sizeof ans);

memset(c,0,sizeof c);

G[1] = 1;

for(int i = 2; i <= 200100;i++){

if(G[i] == -1){

for(int j = i;j <= 200100;j += i){

if(G[j] == -1){

G[j] = i;

}

}

}

}

for(int i=1;i<=200005;i++)

{

Insert(G[i],1);

ans[i] = ans[i-1] + (i-Sum(G[i]));

}

}

int main()

{

init();

int t;

cin>>t;

while(t--)

{

int n;

cin>>n;

cout<

}

return 0;

}

D

解法：模版题，最小覆盖圆

#include

#include

#include

#include

using namespace std;

const double eps=1e-8;

struct Point{

double x,y;

}p[505];

double dis(const Point &a,const Point &b)

{

return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

}

Point circumcenter(const Point &a,const Point &b,const Point &c)

{ //返回三角形的外心

Point ret;

double a1=b.x-a.x,b1=b.y-a.y,c1=(a1*a1+b1*b1)/2;

double a2=c.x-a.x,b2=c.y-a.y,c2=(a2*a2+b2*b2)/2;

double d=a1*b2-a2*b1;

ret.x=a.x+(c1*b2-c2*b1)/d;

ret.y=a.y+(a1*c2-a2*c1)/d;

return ret;

}

void min_cover_circle(Point *p,int n,Point &c,double &r){ //c为圆心，r为半径

random_shuffle(p,p+n); //

c=p[0]; r=0;

for(int i=1;i

{

if(dis(p[i],c)>r+eps) //第一个点

{

c=p[i]; r=0;

for(int j=0;j

if(dis(p[j],c)>r+eps) //第二个点

{

c.x=(p[i].x+p[j].x)/2;

c.y=(p[i].y+p[j].y)/2;

r=dis(p[j],c);

for(int k=0;k

if(dis(p[k],c)>r+eps) //第三个点

{//求外接圆圆心，三点必不共线

c=circumcenter(p[i],p[j],p[k]);

r=dis(p[i],c);

}

}

}

}

}

int main(){

int n;

Point c;

double r;

int t;

cin>>t;

while(t--)

{

cin>>n;

for(int i=0; i

scanf("%lf%lf",&p[i].x,&p[i].y);

min_cover_circle(p,n,c,r);

printf("%.1f %.1f\n",c.x,c.y);

}

return 0;

}

时间： 11-29