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1、http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1106
2、题目大意：
3、题目：
The three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate how much time in a day that all the hands are happy.
Input
The input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1.
Output
For each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places.
Sample Input
0
120
90
-1
Sample Output
100.000
0.000
6.251
#include<stdio.h>
#include<algorithm>
using namespace std;
#define INF 0x7fffffff
double D;
struct node
{double l;double r;
}s[3][2];
node interval(double a,double b)//解方程D<=|a*s-b|<=360-D
{node p;if(a>0){p.l=(D-b)/a;p.r=(360-D-b)/a;}else{p.l=(360-D-b)/a;p.r=(D-b)/a;}if(p.l>=p.r){p.l=0;p.r=0;}if(p.l<0)p.l=0;if(p.r>60)p.r=60;return p;
}node jiao(node a,node b)
{node p;p.l=max(a.l,b.l);p.r=min(a.r,b.r);if(p.l>=p.r) p.l=p.r=0;return p;
}
double solve(int h,int m)
{double a,b;/*时针与分针的角度处理解方程式 D<=|hw-mw|<=360-Dhw=(h+m/60+s/3600)*30mw=(m+s/60)*6*/a=1.0/120.0-1.0/10.0;b=30*h+m/2.0-6.0*m;s[0][0]=interval(a,b);s[0][1]=interval(-a,-b);/*时针与秒针的角度处理解方程式 D<=|hw-sw|<=360-Dhw=(h+m/60+s/3600)*30sw=s*6*/a=1.0/120-6.0;b=30*h+m/2.0;s[1][0]=interval(a,b);s[1][1]=interval(-a,-b);/*分针与秒针的角度处理解方程式 D<=|mw-sw|<=360-Dmw=(m+s/3600)*30sw=s*6*/a=0.1-6;b=6*m;s[2][0]=interval(a,b);s[2][1]=interval(-a,-b);//两个绝对值出来的区间取并集，三个并集之间取交集node s1;double res=0;for(int i=0;i<2;i++)for(int j=0;j<2;j++)for(int k=0;k<2;k++){s1=jiao(jiao(s[0][i],s[1][j]),s[2][k]);res+=s1.r-s1.l;}return res;
}
int main()
{while(scanf("%lf",&D)!=EOF){if(D==-1)break;double ans=0;for(int i=0;i<12;i++){for(int j=0;j<60;j++)ans+=solve(i,j);//i小时j分钟}printf("%.3f\n",100.0*ans/(60*60*12));}return 0;
}
其中的合并取交集参考网上代码，自己写的错了
错的代码：
#include<stdio.h>
#include<algorithm>
using namespace std;
#define INF 0x7fffffff
double D;
struct node
{double l;double r;
}s[3][2];
node interval(double a,double b)//解方程D<=|a*s-b|<=360-D
{node p;if(a>0){p.l=(D-b)/a;p.r=(360-D-b)/a;}else{p.l=(360-D-b)/a;p.r=(D-b)/a;}if(p.l>=p.r){p.l=0;p.r=0;}if(p.l<0)p.l=0;if(p.r>60)p.r=60;return p;
}
node jiao(node a,node b)
{node p;p.l=max(a.l,b.l);p.r=min(a.r,b.r);if(p.l>=p.r) p.l=p.r=0;return p;
}
double solve(int h,int m)
{double a,b;/*时针与分针的角度处理解方程式 D<=|hw-mw|<=360-Dhw=(h+m/60+s/3600)*30mw=(m+s/60)*6*/a=1.0/120-1.0/10.0;b=30*h+m/2.0-6.0*m;s[0][0]=interval(a,b);s[0][1]=interval(-a,-b);/*时针与秒针的角度处理解方程式 D<=|hw-sw|<=360-Dhw=(h+m/60+s/3600)*30sw=s*6*/a=1.0/120-6;b=30*h+m/2.0;s[1][0]=interval(a,b);s[1][1]=interval(-a,-b);/*分针与秒针的角度处理解方程式 D<=|mw-sw|<=360-Dmw=(m+s/3600)*30sw=s*6*/a=0.1-6;b=6.0*m;s[2][0]=interval(a,b);s[2][1]=interval(-a,-b);//两个绝对值出来的区间取并集，三个并集之间取交集node s1[20];int k=0;for(int i=0;i<3;i++){if((s[i][0].r>=s[i][1].l) || (s[i][1].r>=s[i][0].l) ){s1[k].l=max(s[i][0].l,s[i][1].l);s1[k].r=max(s[i][0].r,s[i][1].r);k++;}else if(s[i][0].l>=s[i][1].l && s[i][0].r<=s[i][1].r){s1[k]=s[i][1];k++;}else if(s[i][1].l>=s[i][0].l && s[i][1].r<=s[i][0].r){s1[k]=s[i][0];k++;}else{s1[k]=s[i][0];k++;s1[k]=s[i][1];k++;}}node s2=s1[0];for(int i=1;i<k;i++){s2=jiao(s2,s1[i]);}double anss;anss=s2.r-s2.l;return anss;
}
int main()
{while(scanf("%lf",&D)!=EOF){if(D==-1)break;double ans=0;for(int i=0;i<12;i++){for(int j=0;j<60;j++)ans+=solve(i,j);//i小时j分钟}printf("%.3f\n",100.0*ans/(60*60*12));}return 0;
}
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