(2/5)t i = 1,j = 1; time_t t,u; time(&t); while (i

(2/5)t i = 1,j = 1; time_t t,u; time(&t); while (i sets:shiduan/1..5/:h,t; xinghao/1..3/:NN,m,MM,G,F,Q; link(xinghao,shiduan):n,w; endsets min=@sum(xinghao(i):(G(i)*t(1)+Q(i))*n(i,1)+(w(i,1)-m(i))*n(i,1)*F(i)*t(1))+ @sum(shiduan(j)|j #ge# 2:G(i)*n(i,j)*t(j)+(w(i,j)-m(i))*n(i,j)*F(j)*t(i)+@if(n(i,j)#g int i=2,j=1,k=3 i&&(i+j)&k|i+j matlab多元函数求最小值问题matlab求r =1/(9-j)*(i^2+81-18*j+j^2)^(1/2)*(38/5-19/20*fix(j)+1/20*(8-fix(j))^2)+(fix(j)+1-j)/(9-j)*(i^2+81-18*j+j^2)^(1/2)*(9/5-1/10*fix(j))+(j^2-2*j*n+n^2+m^2-2*m*i+i^2)^(1/2)/(j-n)*(-19/20*fix((n*i-m*j)/(i+n-j- string.Format({0}*{1}={2},i,j,i*j); matlab编程问题,S0=50;K=5;r=0.2;sigma=0.3;T=0.5;a=428.5912b=246.7252c=-801.6449for i=1:10; for j=1:100; A(i,j)=S0*exp((r-sigma^2/2)*T+sigma*random_A(i,j)*sqrt(T)); if A(i,j)>K; and b*A(i,j)+c *(A(i,j)^2)-A(i,j)>-K-a; B(i,j)=0; else B(i,j)=a +b*A(i matlab编程出错,S0=50;K=5;r=0.2;sigma=0.3;T=0.5;a=428.5912b=246.7252c=-801.6449for i=1:10; for j=1:100; A(i,j)=S0*exp((r-sigma^2/2)*T+sigma*random_A(i,j)*sqrt(T)); if A(i,j)>K; and b*A(i,j)+c *(A(i,j)^2)-A(i,j)>-K-a; B(i,j)=0; else B(i,j)=a +b*A(i lingo程序在10行@for(jihe4(i,j)|j#1t#14:x(i,j,3)*x(i,j+1,1)+(1-x(i,j,3))*(1-x(i,j+1,1))=1);错了 求sets:jihe1/1..9/;jihe2/1..14/;jihe3/1..4/;jihe4(jihe1,jihe2);jihe5(jihe1,jihe2,jihe3):x;endsetsmin=@sum(jihe4(i,j):(x(i,j,1)*x(i,j,2)+x(i,j,2)*x(i matlab编程问题 ,不知哪里出错S0=50;K=5;r=0.2;sigma=0.3;T=0.5;q=0;random_A=randn(1000);a=428.5912b=246.7252c=-801.6449for i=1:10; for j=1:100; A(i,j)=S0*exp((r-sigma^2/2)*T+sigma*random_A(i,j)*sqrt(T)); if A(i,j)>K; b*A(i,j)+c *(A(i,j)^2)-A(i int i,j=2;*p=&i;*p=(*&j+1)^j; i=? 用pascal判断三条边可否围成三角形vara:array[1..5,1..3]of integer;n,i,j,m,t:integer;beginwhile not eof dobeginfor t:=1 to 5 do readln(a[t,1],a[t,2],a[t,3]);m:=0;for t:=1 to 5 dobeginn:=a[t,1];i:=a[t,2];j:=a[t,3];if ((n+i)>j)and((n+j)>i)and DIM AS INTEGER A(6),I,T,J,P T=0 FOR I=1 TO 6:A(I)=7-I:NEXT FOR I=1 TO 5 FOR J=1 TO 5 IF A(J)>A( for( int i=0,j=1; j < 5; j+=3 ) i=i+j; 为什么等于5 for(j=i+1;j function f = myfun(x)矩阵定义省略for j=1:28a=0;for i=1;27;a=a+x(i)*A(i,j)*T(i);endJT1(j)=a;endfor j=1:28a=0;for i=1;27;a=a+x(2*i)*A(i,j)*T(i);endJT2(j)=a;endfor j=1:28a=0;for i=1;27;a=a+x(i*3)*A(i,j)*T(i);endJT3(j)=a;endN=[0,0];for i=1:27for j= J = 10 For I = 1 To J Step 5 I = I + 2 J = J - I Next I Print I Print J结果咋是15 和-3呢,要详细的哈,3Q3Q 向量i(i+j)+j(2j+3j)=? 杭电2000题,为什么不能AC#includeusing namespace std;int main( ){char a[5];int i,j;char t;while (cin>>a){for(j=1; j