m=len("1810")m=m+1输出结果是?为什么?

m=len(1810)m=m+1输出结果是?为什么? 下面程序段的输出结果[] m=1; switch (m) { case 1:m++; case 2:m- -; case 3:m++;} printf(“%d”,m); (((m>=1) & (m m*m-5m-1=0则2m*m-5m+1/m*m 已知m*m+m-1=0,求m*m*m+2m*m-2005 这个哪里错了#include #include using namespace std;void strlen(string);void revers(string);void main(){string s,m;int len;cin>>s;coutm;coutlen;strlen(s,len);revers( m,len);}void strlen(string s,int len){char ch[len];for(int i=0;i>ch[i]>>endl;}vo 求Matlab大神,急.bins=[0:L-1];[m,n]=size(x0);len=m*n;x=reshape(x0,len,1);xpdf=hist(x,bins);tr=round(xpdf*triu(ones(L))*(L-1)/len); 求解释上述程序. clear n=0 for i=2 to 1000 k=i*i m=len(alltrim(str(i))) if (k-i)%(10^m)=0 then n=n+1 endif endfor n这个程序中的m=len(alltrim(str(i)))if (k-i)%(10^m)=0 解方程 2m-m*m 1=2(m*m-2m)(m*m-2m) int m=1,n=-1;printf(%d ,(m--&++n));输出结果 m = space(0)c = ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyzfor i = 1 to len(c)nc = asc(substr(c,i,1))if nc>64 and nc86m = m+chr(65+nc-87)ELSEm = m+chr(nc+4)ENDIF else IF nc>120m = m+chr(97+nc-121)ELSEm = m+chr(nc+2)ENDIF endifendfor wait w 化简并求值 m*m*m-2m*m+m除以m*m*m-m 其中m=2011 m^2+m-6/m^2+2m+1÷m+3/m^+m·m^2-1/2-m 其中m=-1/2 n=m/M (m+3m+5m+...+2015m)-(2m+4m+6m+...+2014m)= m-2m-3m+4m-5m-6m+7m.2004m=? (m+3m+5m.+2013m)-(2m+4m+6m+.+2012m)=? m+m*3-m+m=4.8,求m=?