c语言编写查表的程序

用数组来解决最简单方便，自己参考一下吧

#include <stdio.h>

int main()

{

char ch

int i,j,BQ

char * k1[3][4]={

{"0","0.1","0.2-0.3" ,"0.4-0.6"} ,

{"0.1" ,"0.2-0.3","0.4-0.6","0.7-0.9"} ,

{"0.2","0.4-0.6","0.7-0.9","1.0"} ,

}

printf("地下水状态说明:\n")

printf("0-潮湿或点滴状出水\n" )

printf("1-淋雨状或涌流状出水，水压<=0.1MPa或单位出水量<=10L/min\n")

printf("2-淋雨状或涌流状出水，水压>0.1MPa或单位出水量>10L/min\n")

printf("请选择地下水状态[0-2]:")

do {

scanf("%c",&ch )getchar()

} while ( ch >'2' || ch<'0' )

i=ch-'0'

printf("请输入BQ值：")

scanf("%d" , &BQ )

if ( BQ >450 ) j=0 //题目中写的有问题，应该是>450,不是451

else if ( BQ >350 ) j=1

else if ( BQ >250 ) j=2

else j=3

printf("K1=%s\n" , k1[i][j] )

system("pause")

return 0

}

看着有点模糊，两者之间是什么关系，当温度/摄氏度0度时是溶解度/克 29.4，当温度/摄氏度 80，溶解度/克 65.6

是的话(温度/摄氏度设一个变量T,溶解度/克设一个数组变量R[9] )

#include<stdio.h>

main()

{

int T=0,i

float R[9]=｛29.4,33.3,37.2,41.4,45.8,50.4,55.2,60.2,65.6｝

printf("Please enter temperature /Degrees Celsius \n ")

printf("between 0 to 80 and is a multiple of 10 \n")

scanf("%d",&T)

for(i=0i<=80i+10)//表格温度/摄氏度是否在0~80且是10的倍数

{

if(T=i)//温度是否没有输错

printf("%d,%f",T,R[i/10])//找到的话输出相应的温度/摄氏度与溶解度/克

else

printf(" Enter the temperature can't match ")//输的温度与/摄氏度有错

}

}

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