度数序列Problem:ATimeLimit:1000msMemoryLimit:65536KDescription由握手定理我们懂得任意的一种图中，所有顶点的度数之和等于边数的2倍，那么给你一种无向图的度数序列，你能鉴定它能否构成无向图吗？(10分）Input输入数据有多组，每组第一行有1个数为n(1<=n<=100)，接下来第二行有n个正整数，代表n个度数。Output假如能构成图，则在一行内输出yes，否则输出no。SampleInput41234SampleOutputYes#include
#includeusingnamespacestd;intmain(){intn;while(scanf("%d",&n)!=-1){ints=0;for(inti=0;i>a;s+=a;}intb=(s&1);if(b==1){cout<<"no"<#includeusingnamespacestd;intmain(){intx,y;while(~scanf("%d%d",&x,&y)){cout<<(x+y-2)<#includeusingnamespacestd;intmain(){intn;while(scanf("%d",&n)!=-1){cout<<2*(n-1)<#includeusingnamespacestd;intD(intx){if(x==1){return0;}if(x==2){return1;}return(x-1)*(D(x-2)+D(x-1));}intmain(){intn;while(scanf("%d",&n)!=-1){cout<#includeusingnamespacestd;longa[100];intf(intn){if(n==1){return7;}return6*f(n-1)+a[n-1];}intmain(){a[0]=1;for(inti=1;i<20;i++){a[i]=a[i-1]*8;}intn;while(scanf("%d",&n)!=-1){cout<#includeusingnamespacestd;longa[100];longb[100];intmain(){a[0]=1;b[0]=1;for(inti=1;i<20;i++){a[i]=a[i-1]*2;}intn;while(scanf("%d",&n)!=-1){if(n==0){cout<<1<#includeusingnamespacestd;intmain(){intn,m;while(scanf("%d%d",&n,&m)!=-1){...
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