巧妙的方法.doc

统计问题Description给出一个带n个顶点的凸多边形,我们保证它不存在3条对角线相交于同一个点。请统计每两条对角线的交点数的和。n=6时,图中的15个圆点即为所求交点Input输入只含一个整数n(3≤n≤100)。Output输出交点数。SampleInput6SampleOutput15Hint一个多边形是凸多边形当且仅当每个内角都小于180°。代码:#include<>intmain(){intn;while(scanf("%d",&n)!=EOF){intallb;allb=n*(n-1)*(n-2)*(n-3)/24;printf("%d\n",allb);}return0;}1341巧用异或WhowillbepunishedDescriptionThistime,suddenly,,,Lidoesn',thistime,,but,itistroublesome,and,teacheralwayshavemanythingstothinkabout,so,teacherLiwantsyou,whoisintheACMteam,,andNwillnotgreaterthan500,,followingNlines,--etoclassthistime,',firstprintalinesaying"Scenario#k",,#1A代码A:#include<iostream>#include<cstdio>#include<string>usingnamespacestd;intmain(){intt;intk=1;while(cin>>t){stringstr,sum;cin>>sum;for(intj=1;j<2*t-1;j++){cin>>str;for(inti=0;i<();i++)sum[i]=sum[i]^str[i];}printf("Scenario#%d\n",k);cout<<sum<<endl<<endl;k++;}return0;}代码B:#include<iostream>#include<cstdio>#include<string>usingnamespacestd;intmain(){intt;intk=1;intsum[32];while(cin>>t){stringstr;for(inti=0;i<32;i++)sum[i]=0;for(intj=1;j<=t;j++){cin>>str;for(inti=0;i<();i++)sum[i]=sum[i]+str[i];}for(intj=1;j<=t-1;j++){cin>>str;for(inti=0;i<();i++)sum[i]=sum[i]-str[i];}printf("Scenario#%d\n",k);for(inti=0;sum[i