杭電C語(yǔ)言課程設(shè)計(jì)實(shí)驗(yàn)六答案

版權(quán)：ydz題：1．使用函數(shù)指針編寫用梯形公式計(jì)算定積分和的值，取。提示：將積分區(qū)間[a,b]作n等分：，，，在每一分段上取點(diǎn)，，每一小段上用梯形來(lái)近似代替曲邊梯形得：記上式右邊為，則：2．使用數(shù)組指針處理方陣，方陣的階數(shù)定義為常量N。用數(shù)組指針作函數(shù)參數(shù)完成下列功能，每個(gè)功能編一個(gè)函數(shù)。實(shí)現(xiàn)方陣的初始化；實(shí)現(xiàn)方陣的輸出；實(shí)現(xiàn)兩個(gè)方陣相加，并輸出相加結(jié)果；實(shí)現(xiàn)兩個(gè)方陣相減，并輸出相減結(jié)果；實(shí)現(xiàn)兩個(gè)方陣相乘，并輸出相乘結(jié)果；主函數(shù)定義兩個(gè)整型二維數(shù)組，分別調(diào)用上述函數(shù)去實(shí)現(xiàn)功能。程序一#include<stdio.h>#include<math.h>double(*f)(doublex);doublet1,t2;doublejifen(doublea,doubleb,double(*f)(doublex)){ doublet=0,h,x; inti,n=100000; h=(b-a)/n; x=a; for(i=1;i<=n;i++){ t=t+0.5*h*f(x); x=a+i*h; } returnt;}doubleg(doublex){ returnsin(x)/x;}doublep(doublex){ return1/(1+x*x);}voidmain(){ t1=jifen(1,2,g); t2=jifen(0,1,p); printf("%lf\n%lf\n",t1,t2);}程序二#include<stdio.h>#defineN5voidchushihua(int(*p)[N]){ inti,j; for(i=0;i<N;i++){ for(j=0;j<N;j++){ scanf("%d",p[i]+j); } }}voidshuchu(int(*p)[N]){ inti,j; for(i=0;i<N;i++){ for(j=0;j<N;j++){ printf("%d",*(p[i]+j)); } printf("\n"); }}voidxiangjia(int(*p)[N],int(*q)[N],int(*r)[N]){ inti,j; for(i=0;i<N;i++){ for(j=0;j<N;j++){ *(r[i]+j)=*(p[i]+j)+*(q[i]+j); printf("%d",*(r[i]+j)); } printf("\n"); }}voidxiangjian(int(*p)[N],int(*q)[N],int(*r)[N]){ inti,j; for(i=0;i<N;i++){ for(j=0;j<N;j++){ *(r[i]+j)=*(p[i]+j)-*(q[i]+j); printf("%d",*(r[i]+j)); } printf("\n"); }}voidxiangcheng(int(*p)[N],int(*q)[N],int(*r)[N]){ inti,j,k; fo