2012 金华 现场赛

A 水题

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define mod 31536000#define eps 1e-10typedef __int64 ll;struct node{    int a,b;    double id;}M[500010];int cmp(node a,node b){    return (long long)a.a*b.b<(long long)b.a*a.b;}long long sum,time;int main(){    int i,j,k,n;    while(scanf("%d",&n)!=EOF)    {        if(n==0) break;        for(i=0;i<n;i++)        {            scanf("%d%d",&M[i].a,&M[i].b);        }        sort(M,M+n,cmp);        sum=0;time=0;        for(i=0;i<n;i++)        {            if(M[i].a==0) continue;            sum+=(long long)(M[i].a+(long long)time*M[i].b);            sum%=mod;            time+=(long long)(M[i].a+(long long)time*M[i].b);            time%=mod;        }       //rintf("%lld %lld",time,sum);        printf("%lld\n",sum%mod);    }    return 0;}

D 枚举每一个角度，不知道是不是数据水，枚举1000次就可以了

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const double PI=acos(-1.0);const double g = 9.8;double H,l1,l2,r1,r2;int n;double a[300];int cal(double x){    int ret = 0;    for(int i = 0;i<n;i++){        double vx = a[i]*cos(x);        double vy = a[i]*sin(x);        double V = sqrt(2.0*g*H+vy*vy);        double t = (V-vy)/g;        double d = 1.0*t*vx;        if(d<=r2&&d>=l2) return 0;        if(d<=r1&&d>=l1) ret++;    }    return ret;}int main(){        while(scanf("%d",&n)&&n!=0){            scanf("%lf%lf%lf%lf%lf",&H,&l1,&r1,&l2,&r2);            for(int i=0;i<n;i++){                scanf("%lf",&a[i]);            }            double add = PI/1000;            int ans = 0;            for(double i=-PI/2;i<=PI/2;i+=add){                ans = max(ans,cal(i));            }            printf("%d\n",ans);        }}

I 签到题 所有的平方之和

#include <cstdio>int a[1000];int main(){   int n;   while(scanf("%d",&n)&&n!=0){     int ans = 0;    for(int  i = 0;i<n;i++){        scanf("%d",&a[i]);        ans += a[i]*a[i];    }    printf("%d\n",ans);   }}

J 题意没有看 直接样例脑补了一下

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,m,k;int p;int vis1[2222];int vis2[2222];char str1[20];char str2[20];int main(){   while(scanf("%d%d%d",&n,&m,&k)&&n+m+k!=0){      scanf("%d",&p);      int ans = 0;      memset(vis1,0,sizeof(vis1));      memset(vis2,0,sizeof(vis2));      int a,b;    while(p--){        scanf("%s%d%s%d",str1,&a,str2,&b);        if(str1[0]=='c') {vis1[b]++;ans+=k;}        else {vis2[a]++,ans+=n;}    }    for(int i=1;i<=m;i++){        ans-=vis1[i]*vis2[i];    }    printf("%d\n",n*m*k-ans);   }}

K 题意开始看错了 ，应该是在每一次时间结束才转弯，中间过程碰到是不转弯的

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n;int time;struct Point{int dir;int x,y;int s,t;void go_next()    {if(dir==0){x-=s;if(x<1) {x=2-x;dir=2;}}else if(dir==1){y-=s;if(y<1) {y=2-y;dir=3;}}else if(dir==2){x+=s;if(x>n) {x=2*n-x;dir=0;}}else{y+=s;if(y>n) {y=n*2-y;dir=1;}}}}T,J;int get_dir(char c){if(c=='N') return 0;else if(c=='W') return 1;else if(c=='S') return 2;else return 3;}int main(){while(scanf("%d",&n)&&n!=0){T.x=T.y=1;J.x=J.y=n;char op[10];scanf("%s%d%d",op,&T.s,&T.t);T.dir=get_dir(op[0]);scanf("%s%d%d",op,&J.s,&J.t);J.dir=get_dir(op[0]);scanf("%d",&time);for(int i=1;i<=time;i++){T.go_next();J.go_next();if(T.x==J.x&&T.y==J.y){swap(T.dir,J.dir);continue;}if(i%T.t==0) T.dir=(T.dir+1)%4;if(i%J.t==0) J.dir=(J.dir+1)%4;}printf("%d %d\n",T.x,T.y);        printf("%d %d\n",J.x,J.y);}return 0;}