本文转载自博客
http://blog.csdn.net/mzx0821
雨纷纷、旧故里草木深、我听闻、你始终一个人、斑驳的城门、盘踞着老树根、石板上回荡的是再等、
—— 永不放弃的Mzx0821
希望全部的同学算法分析都都不会挂科、
注：题库包含卓越班实验题和非卓越班实验题、由于不知道老师会不会考超范围的、
由于OJ数据弱的原因、不保证下面解法的正确性、特别感谢small rabbit贡献的部分答案、
非卓越班实验题汇总
实验一：
249 凸包面积
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct P{
double x,y;
P(){}
P(double x1,double y1)
{
x=x1;
y=y1;
}
double add(double a,double b)
{
return a+b;
}
P operator + (P p)
{
return P(add(x,p.x),add(y,p.y));
}
P operator - (P p)
{
return P(add(x,-p.x),add(y,-p.y));
}
double det(P p)
{
return x*p.y-y*p.x;
}
};
bool cmp(P a,P b)
{
if(a.x != b.x)
return a.x<b.x;
return a.y<b.y;
}
P p[1005],s[1005];
int main()
{
int n;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int i;
for(i=0;i<n;++i)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
int k=0;
sort(p,p+n,cmp);
for (i=0;i<n;++i)
{
while(k>=2 && (s[k-1]-s[k-2]).det(p[i]-s[k-2]) <= 0)
k--;
s[k++]=p[i];
}
int t=k;
for(i=n-1;i>=0;i--)
{
while(k>t && (s[k-1]-s[k-2]).det(p[i]-s[k-2]) <= 0)
k--;
s[k++]=p[i];
}
double ans=0;
while(k>=2)
{
P p1=s[k-1];
P p2=s[k];
P p3=s[0];
ans+=fabs(0.5*(p1.det(p2)+p2.det(p3)+p3.det(p1)));
k--;
}
printf("%.1lf
",ans);
}
return 0;
}303 取模
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
#define LL long long
struct MOD
{
LL init(char str[],LL z)
{
LL temp=0;
for(int i=0;str[i];i++)
{
temp=temp*10+str[i]-'0';
temp%=z;
}
return temp;
}
LL solve(LL x,LL y,LL z)
{
LL ans=1;
while(y)
{
if(y&1)ans=ans*x%z;
x=x*x%z;
y>>=1;
}
return ans;
}
};
int main()
{
MOD mod;
char str[105];
LL x,y,z;
while(scanf("%s%lld%lld",str,&y,&z)!=EOF)
{
x=mod.init(str,z);
printf("%lld
",mod.solve(x,y,z));
}
}352 合并果子
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
class MergeFruit
{
private:
priority_queue<int, vector<int>, greater<int> >q;
int n;
int x;
int sum;
public:
void Init()
{
sum=0;
}
void Scan()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
q.push(x);
}
}
void Run()
{
while(!q.empty())
{
int min_1=q.top();
q.pop();
if(q.empty())break;
int min_2=q.top();
q.pop();
int ans=min_1+min_2;
q.push(ans);
sum+=ans;
}
}
void Print()
{
cout<<sum<<endl;
}
};
int main()
{
MergeFruit mergefruit;
mergefruit.Init();
mergefruit.Scan();
mergefruit.Run();
mergefruit.Print();
return 0;
}493 PostOffice
#include <stdio.h>
#include <math.h>
#include<algorithm>
using namespace std;
int main()
{
int n,i,mid,length=0;
int pos_x[10005],pos_y[10005];
scanf("%d",&n);
for (i=0;i<n;i++)
scanf("%d %d",&pos_x[i],&pos_y[i]);
sort(pos_x,pos_x+n);
sort(pos_y,pos_y+n);
mid=n/2;
for(i=0;i<n;i++)
length+=abs(pos_x[i]-pos_x[mid])+abs(pos_y[i]-pos_y[mid]);
printf("%d
",length);
return 0;
}794 近期对问题
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define inf 0x7ffffff
struct node{
double x;
double y;
}a[100001];
int temp[100001];
bool cmp(node a,node b)
{
if(a.x != b.x)
return a.x<b.x;
return a.y<b.y;
}
bool cmp1(int c,int b)
{
return a[c].y<a[b].y;
}
double min(double a,double b)
{
return a<b?a:b;
}
double dis(int i,int j)
{
return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)
+(a[i].y-a[j].y)*(a[i].y-a[j].y));
}
double Find(int l,int r)
{
double d;
if(l == r)
return inf;
if(l + 1 == r)
return dis(l,r);
int mid=(l+r)>>1;
double d1=Find(l,mid);
double d2=Find(mid+1,r);
d=min(d1,d2);
int cnt=0;
int i,j;
for(i=l;i<=r;++i)
{
if(fabs(a[i].x-a[mid].x) <= d)
temp[cnt++]=i;
}
sort(temp,temp+cnt,cmp1);
for(i=0;i<cnt;++i)
{
for(j=i+1;j<cnt && (a[temp[j]].y-a[temp[i]].y < d);++j)
{
d=min(d,dis(temp[i],temp[j]));
}
}
return d;
}
int main()
{
int T;
int n;
scanf("%d",&T);
while(T--)
{
int i,j;
scanf("%d",&n) ;
for(i=0;i<n;++i)scanf("%lf%lf",&a[i].x,&a[i].y);
sort(a,a+n,cmp);
printf("%.4lf
",Find(0,n-1));
}
return 0;
}实验二：76 数字模式的识别
#include<stdio.h>
#define N 4000001
int a[N]={0};
int main()
{
int n,i,b,x,y;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&b);
a[b+2000000]++;
}
x=0;
for(i=0;i<N;i++)
{
if(a[i]>x)
{
x=a[i];
y=i;
}
}
printf("%d
",y-2000000);
return 0;
}254 翻煎饼
#include<iostream>
using namespace std;
int s[1000];
int main()
{
int i,n,k=0;
cin>>n;
for(i=0l;i<n;i++)
cin>>s[i];
for(;n>1;n--)
{
int max=0,l=0;
for(i=0;i<n;i++)
{
if(s[i]>max)
{
max=s[i];
l=i;
}
}
if(l==0)
{
k+=1;
for(i=0;i<n/2;i++)
{
int
t=s[i];
s[i]=s[n-1-i];
s[n-1-i]=t;
}
}
else if(l<n-1)
{
k+=2;
for(i=0;i<(l+1)/2;i++)
{
int t=s[i];
s[i]=s[l-i];
s[l-i]=t;
}
for(i=0;i<n/2;i++)
{
int t=s[i];
s[i]=s[n-1-i];
s[n-1-i]=t;
}
}
}
cout<<k<<endl;
}342 变位词
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
char s1[1000],s2[1000];
int n,len1,len2;
cin>>n;
while(n--)
{
cin>>s1>>s2;
len1=strlen(s1);
len2=strlen(s2);
sort(s1,s1+len1);
sort(s2,s2+len2);
if(strcmp(s1,s2)==0)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
return 0;
}445 选择问题
#include<iostream>
using namespace std;
int main()
{
int n,i,j,a[10002],temp,m;
cin >> n >> m;
for(i=0;i<n;i++)
{
cin >> a[i];
}
for(i=0;i<n;i++)
{
for(j=0;j<n-i-1;j++)
{
if(a[j]>a[j+1])
{
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
}
cout << a[m-1] << endl;
return 0;
}541 排列的字典序问题
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
int a[55];
int t[55];
int main()
{
int n;
ll num[15];
num[1]=1;
int i;
for(i=2;i<15;++i)
num[i]=num[i-1]*i;
while(scanf("%d",&n) != EOF)
{
ll ans=0;
int ss=0;
int f=-1;
for(i=0;i<n;++i)scanf("%d",&a[i]);
for(i=0;i<n;++i)
{
int count=0;
int j;
for(j=i+1;j<n;++j)
{
if(a[i]>a[j])
count++;
}
t[i]=count;
}
for(i=0;i<n;++i)
{
ans+=t[i]*num[n-1-i];
}
for(i=n-1;i>=0;--i)
{
if(a[i]<ss && f == -1)
{
f=i;
}
ss=a[i];
}
int tt=a[f];
int vis;
sort(a+f,a+n);
for(i=f;i<n;++i)
{
if(a[i]>tt)
{
vis=i;
break;
}
}
printf("%lld
",ans);
for(i=0;i<n;++i)
{
if(i == vis)
continue;
if(i == f)
printf("%d ",a[vis]);
printf("%d ",a[i]);
}
printf("
");
}
return 0;
}642 俄式乘法
#include<cstdio>
#include<cstring>
int main()
{
int n,m;
while(scanf("%d%d",&n,&m) != EOF)
{
bool first=0;
int ans=0;
while(n)
{
if(n%2 == 0)
{
n/=2;
m*=2;
}
else
{
if(first)
printf(" + ");
n=(n-1)/2;
printf("%d",m);
first=1;
ans+=m;
m=m*2;
}
}
printf(" = %d
",ans);
}
return 0;
}实验三：
传送门：http://blog.csdn.net/mzx0821/article/details/41593241
卓越班实验题目汇总
#include<cstdio>
#include<cstring>
int a[5000001];
int main()
{
int n,k;
while(scanf("%d%d",&n,&k) != EOF)
{
int i;
for(i=0;i<n;++i)
{
scanf("%d",&a[i]);
}
int ans=-1;
int l=0,r=n-1;
while(l <= r)
{
int mid=(l+r)/2;
if(a[mid] == k)
{
ans=mid;
break;
}
if(a[mid] > k)
r=mid-1;
else
l=mid+1;
}
printf("%d
",ans+1);
}
return 0;
}956 约瑟夫问题的实现#include<cstdio>
int main()
{
int n,m;
while(scanf("%d%d",&n,&m) != EOF)
{
int ans=0;
for(int i=2;i<=n;++i)
ans=(ans+m)%i;
printf("%d",ans+1);
}
return 0;
}
#include<cstdio>
int a,b;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
int main()
{
while(scanf("%d%d",&a,&b) != EOF)
{
int t=gcd(a,b);
a/=t;
b/=t;
if(a%2 == 1)printf("A
");
else printf("B
");
}
return 0;
}405 Fibonacci number
#include<stdio.h>
int a[41];
int main()
{
int n,i;
scanf("%d",&n);
a[0]=0;
a[1]=1;
a[2]=1;
for(i=3;i<=n;i++)
{
a[i]=a[i-1]+a[i-2];
}
printf("%d
",a[n]);
return 0;
}413 Quick Sort#include<cstdio>
#include<cstring>
int a[50005];
void quick_sort(int l,int r,int a[])
{
if(l<r)
{
int i=l,j=r;
int temp;
temp=a[l];
while(i<j)
{
while(i<j && a[j] >= temp)
j--;
if(i<j)
a[i++]=a[j];
while(i<j && a[i] <= temp)
i++;
if(i<j)
a[j--]=a[i];
}
a[i]=temp;
quick_sort(l,i-1,a);
quick_sort(i+1,r,a);
}
}
int main()
{
int n;
while(scanf("%d",&n) != EOF)
{
for(int i=0;i<n;++i)
{
scanf("%d",&a[i]);
}
quick_sort(0,n-1,a);
for(int i=0;i<n;++i)
{
printf("%d ",a[i]);
}
printf("
");
}
return 0;
}414 The Next Permutation#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
char str[1005];
int main()
{
int m,n,len;
cin>>m;
memset(str,0,sizeof(str));
while(m--)
{
cin>>n>>str;
len=strlen(str);
if(next_permutation(str,str+len))
cout<<n<<" "<<str<<endl;
else
cout<<n<<" "<<"BIGGEST"<<endl;
}
return 0;
}425 Polynomial calculate#include<cstdio>
#include<cstring>
int main()
{
int n,x;
while(scanf("%d%d",&n,&x) != EOF)
{
int a[22];
for(int i=0;i<=n;++i)
{
scanf("%d",&a[i]);
}
if(n == 0)
{
printf("0
");
continue;
}
int sum=a[n];
for(int i=n-1;i>=0;i--)
sum=sum*x+a[i];
printf("%d
",sum);
}
return 0;
}446 合并排序#include<cstdio>
int a[10005];
int b[10005];
void merge(int a[],int l,int mid,int r)
{
int cnt=0;
int i=l;
int j=mid+1;
while(i<=mid && j <= r)
{
if(a[i] < a[j])
{
b[cnt++]=a[i++];
}
else
b[cnt++]=a[j++];
}
while(i<=mid)
b[cnt++]=a[i++];
while(j<=r)
b[cnt++]=a[j++];
int t=0;
for(int i=l;i<=r;++i)
a[i]=b[t++];
}
void hebing(int l,int r,int a[])
{
int mid=(l+r)/2;
if(l<r)
{
hebing(l,mid,a);
hebing(mid+1,r,a);
merge(a,l,mid,r);
}
}
int main()
{
int n;
while(scanf("%d",&n) != EOF)
{
for(int i=0;i<n;++i)
scanf("%d",&a[i]);
hebing(0,n-1,a);
for(int i=0;i<n-1;++i)
{
if(i%10 == 0)
printf("%d",a[i]);
else if((i+1)%10 == 0)
printf(" %d
",a[i]);
else
printf(" %d",a[i]);
}
if((n-1)%10 == 0)
printf("%d
",a[n-1]);
else
printf(" %d
",a[n-1]);
}
return 0;
} 480 Locker doors#include<cstdio>
#include<cstring>
#include<cmath>
int a[100005];
int main()
{
for(int i=1;i <= 100000;++i)
{
int t=sqrt((double)i);
if(t*t == i)
a[i]=1;
a[i]+=a[i-1];
}
int n;
while(scanf("%d",&n) != EOF)
{
printf("%d
",a[n]);
}
return 0;
}641 The Dutch flag problem#include<cstdio>
char str[500005];
int main()
{
int numR,numB,numW;
int n;
while(scanf("%d",&n) != EOF)
{
numR=numB=numW=0;
scanf("%s",str);
for(int i=0;str[i];++i)
{
if(str[i] == 'W')
numW++;
else if(str[i] == 'B')
numB++;
else if(str[i] == 'R')
numR++;
}
for(int i=0;i<numR;++i)
printf("R");
for(int i=0;i<numW;++i)
printf("W");
for(int i=0;i<numB;++i)
printf("B");
printf("
");
}
return 0;
}411 售货员的难题#include<cstdio>
#include<cstring>
#define inf 0x3f3f3f3f
int dp[21][1<<20];
int map[25][25];
int n;
int min(int a,int b)
{
return a<b?a:b;
}
int solve()
{
int i,j,k;
int st=1<<(n+1);
for(i=0;i<=n;++i)
for(j=0;j<st;++j)
dp[i][j]=inf;
dp[0][1]=0;
for(i=1;i<st;++i)
{
for(j=0;j<=n;++j)
{
if((i&(1<<j)) == 0)
continue;
for(k=0;k<=n;++k)
{
if((i&(1<<k)) == 0)
dp[k][i|(1<<k)]=min(dp[k][i|(1<<k)],dp[j][i]+map[j][k]);
}
}
}
return dp[n][st-1];
}
int main()
{
while(scanf("%d",&n) != EOF)
{
int i,j;
for(i=0;i<n;++i)
{
for(j=0;j<n;++j)
scanf("%d",&map[i][j]);
map[i][n]=map[i][0];
}
int ans=solve();
printf("%d
",ans);
}
return 0;
}
572 Boyer–Moore–Horspool algorithm#include<cstdio>
#include<cstring>
char str1[200005];
char str2[800005];
int next[800005];
void getnext(int len)
{
int j,k;
j=0,k=-1;
next[0]=-1;
while(j<len-1)
{
if(k == -1 || str1[j] == str1[k])
{
j++;
k++;
next[j]=k;
}
else
k=next[k];
}
}
int kmpindex(int len1,int len2)
{
int i=0,j=0;
getnext(len2);
while(j<len1 && i<len2)
{
if(j == -1 || str1[j] == str2[i])
{
i++;
j++;
}
else
j=next[j];
}
if(j>=len1)
return i-len1;
return -1;
}
int main()
{
while(scanf("%s%s",str1,str2) != EOF)
{
int len1=strlen(str1);
int len2=strlen(str2);
int ans=kmpindex(len1,len2);
printf("%d
",ans);
}
return 0;
}649 NBA Finals#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
float dp[505][505];
int main()
{
int n;
float p,q;
while(cin>>n>>p)
{
q=1-p;
memset(dp,0,sizeof(dp));
float ans=0;
for(int i=0;i<=n;i++)
dp[0][i]=1;
for(int i=1;i<=n;++i)
{
for(int j=1;j<=n;++j)
{
dp[i][j]=dp[i-1][j]*p+dp[i][j-1]*q;
}
}
cout<<dp[n/2+1][n/2+1]<<endl;
}
return 0;
}680 Jack Straws
#include <iostream> #include <memory> #define MAX_N 13 #define MAXV(x, y) ((x) >= (y) ?410 尼克的任务(x) : (y)) #define MINV(x, y) ((x) <= (y) ? (x) : (y)) using namespace std; int num; struct node { int x1, y1, x2, y2; }data[MAX_N + 1]; int set[MAX_N + 1]; int rank[MAX_N + 1]; //推断第3个点在1,2点构成的线短的哪个方向, -1: 逆时针方向, 1: 顺时针方向 int direct(int x1, int y1, int x2, int y2, int x3, int y3) { return (x3 - x1) * (y2 - y1) - (x2 - x1) * (y3 - y1); } //推断第3个点是否在1,2点构成的线短上 bool onLine(int x1, int y1, int x2, int y2, int x3, int y3) { return (MINV(x1, x2) <= x3 && x3 <= MAXV(x1, x2) && MINV(y1, y2) <= y3 && y3 <= MAXV(y1, y2)); } //推断点1,2构成的线短与点3,4构成的线短是否相交 bool intersect(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) { int d1 = direct(x3, y3, x4, y4, x1, y1); int d2 = direct(x3, y3, x4, y4, x2, y2); int d3 = direct(x1, y1, x2, y2, x3, y3); int d4 = direct(x1, y1, x2, y2, x4, y4); if(((d1 > 0 && d2 < 0) || (d1 < 0 && d2 > 0)) && ((d3 > 0 && d4 < 0) || (d3 < 0 && d4 > 0))) return true; if(d1 == 0 && onLine(x3, y3, x4, y4, x1, y1)) return true; else if(d2 == 0 && onLine(x3, y3, x4, y4, x2, y2)) return true; else if(d3 == 0 && onLine(x1, y1, x2, y2, x3, y3)) return true; else if(d4 == 0 && onLine(x1, y1, x2, y2, x4, y4)) return true; else return false; } int find(int pos) { if(pos != set[pos]) set[pos] = find(set[pos]); return set[pos]; } void joinSet(int pos1, int pos2) { if(pos1 == pos2) return; int pre1 = find(pos1); int pre2 = find(pos2); if(pre1 == pre2) return; else { int rank1 = rank[pre1]; int rank2 = rank[pre2]; if(rank1 < rank2) set[pre1] = pre2; else if(rank1 > rank2) set[pre2] = pre1; else { rank[pre1]++; set[pre2] = pre1; } } } void init() { int i; for(i = 1; i <= num; i++) { set[i] = i; rank[i] = 0; } } int main() { int i, n1, n2, pre1, pre2; while(cin>>num && num != 0) { for(i = 1; i <= num; i++) cin>>data[i].x1>>data[i].y1>>data[i].x2>>data[i].y2; init(); for(n1 = 1; n1 <= num; n1++) { for(n2 = n1; n2 <= num; n2++) { pre1 = find(n1); pre2 = find(n2); if(pre1 == pre2) continue; else { if(intersect(data[n1].x1, data[n1].y1, data[n1].x2, data[n1].y2, data[n2].x1, data[n2].y1, data[n2].x2, data[n2].y2)) joinSet(n1, n2); } } } while(cin>>n1>>n2 && !(n1 == 0 && n2 == 0)) { pre1 = find(n1); pre2 = find(n2); if(pre1 == pre2) cout<<"CONNECTED"<<endl; else cout<<"NOT CONNECTED"<<endl; } } return 0; }
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int p[10005][20], dp[10005];
int Time, N;
int s, len;
while(cin>>Time>>N)
{
memset(p, 0, sizeof(p));
memset(dp, 0, sizeof(dp));
for (int i = 0; i < N; i++)
{
cin>>s>>len;
p[s][0]++;
p[s][p[s][0]] = len;
}
for (int i = Time; i >= 1; i--)
{
if (p[i][0] == 0)
{
dp[i] = dp[i + 1] + 1;
}
else
if (p[i][0] == 1)
{
dp[i] = max(dp[i + p[i][1]], dp[i]);
}
else
{
for (int j = 1; j <= p[i][0]; j++)
{
dp[i] = max(dp[i], dp[i + p[i][j]]);
}
}
}
cout<<dp[1]<<endl;
}
return 0;
}544 跑跑卡丁车#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define INF 0x73737373
int a[10010], b[10010], dp[10010][15];
void input(int L, int N, int a[])
{
for(int i = 1; i <= L; i++)
{
scanf("%d", &a[i]);
for(int j = 1; j < N; j++)
a[i + j * L] = a[i];
}
}
int main()
{
int L, N;
while(~scanf("%d%d", &L, &N))
{
input(L, N, a);
input(L, N, b);
for(int i = 0; i <= L * N; i++)
for(int j = 0; j < 15; j++)
dp[i][j] = INF;
dp[0][0] = 0;
for(int i = 1; i <= L * N; i++)
{
for(int j = 0; j < 15; j++)
{
if(j != 0)dp[i][j] = min(dp[i][j], dp[i-1][j-1] + a[i]);
if(j == 10) dp[i][j] = min(dp[i][j], dp[i-1][14] + a[i]);
if(j < 10)dp[i][j] = min(dp[i][j], dp[i-1][j+5] + b[i]);
}
}
int ret = INF;
for(int i = 0; i < 15; i++)
ret = min(dp[L * N][i], ret);
printf("%d
", ret);
}
return 0;
}679 Secret Code#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
const int maxn=110;
LL xr,xi,br,bi,num;
LL flag,t;
LL ans[maxn];//保存枚举的ai
void dfs(LL rr,LL ii,LL step)
{
LL x,y,i;
if (step>100)return;
if(flag)return;
if(rr==0&&ii==0)
{
flag=1;
t=step;
return;
}
for(i=0;i*i<num;i++)
{
//复数除法运算
x=(rr-i)*br+ii*bi;
y=ii*br-(rr-i)*bi;
ans[step]=i;
if(x%num==0&&y%num==0)//保证整除
dfs(x/num,y/num,step+1);
if(flag)return;
}
}
int main()
{
LL T,i;
scanf("%lld",&T);
while(T--)
{
scanf("%lld%lld%lld%lld",&xr,&xi,&br,&bi);
num=br*br+bi*bi;
flag=0;
dfs(xr,xi,0);
if(!flag)printf("The code cannot be decrypted.
");
else
{
printf("%lld",ans[t-1]);
for(i=t-2;i>=0;i--)
printf(",%lld",ans[i]);
printf("
");
}
}
return 0;
}
698 Independent Task Scheduling
#include <stdio.h>
int main()
{
float x1,x2,x3,y1,y2,y3;
float s;
while(scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF)
{
s=(x1*y2+x2*y3+x3*y1-x3*y2-x1*y3-x2*y1)/2.0;
if(s<0)s=-s;
printf("%.1f
",s);
}
return 0;
}1080 单纯行法。
。数学渣、不会做、等有空问下理学院的同学怎么搞、