东林离散大作业

度数序列Problem:ATimeLimit:1000msMemoryLimit:65536KDescription由握手定理我们知道任意的一个图中，所有顶点的度数之和等于边数的 2 倍，那么给你一个无向图的度数序列，你能判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如果能构成图，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<