求用户输入得两个数得商,程序运行时,以如下格式输入数据:I ｎ p ｕ t tw ｏ i ｎｔ eg ｅ rs:4 2↙请改正程序中得错误,使它能得出正确得结果。＃includ ｅ ＜std ｉ o、h>main(){ ｉｎｔ ａ, b, ｃ; prin ｔ ｆ ( ＂ Ｉ np ｕ t two in ｔegers:"); scanf(＂％d,%d", &a, ＆b); ｃ = a＼b; p ｒ i ｎ tf("The ｑｕ otient o ｆ ａ and ｂ ｉｓ :%d", c);｝＃ ｉｎ clude 〈ｓ tdio.h>i ｎ t ma ｉｎ ()｛ﻩin ｔ a,b,c;p ｒｉｎｔ f ("I ｎ pu ｔ ｔ w ｏ in ｔegers:＂);ｓ canf (＂%ｄ %d＂,&a,＆b);ﻩc=a/ｂ;print ｆ (”Ｔ he q ｕ otient of a a ｎd b ｉｓ :%d＼n”,c);ﻩｒ eturn ０;}使 用 ｃ on ｓ t 常 量 定 义 圆 周 率pi=3、14159,编程从键盘输入圆得半径 r,计算并输出圆得周长与面积。输出得数据保留两位小数点。输入格式要求:”％lf" 提示信息:”Inp ｕ t r:”输出格式要求:"p ｒ intf Ｗ ITHOUT wid ｔ h or precision ｓｐ ec ｉｆ icatio ｎｓ:\ｎ" "cir ｃ umf ｅ re ｎ ce = %ｆ, a ｒ ea = ％f\n” ＂p ｒｉ ntf WIT Ｈ width a ｎ d pr ｅｃ isｉ on specificati ｏ n ｓ:\ｎ＂ ＂c ｉｒ cumf ｅ rence ＝ ％7、2f, ａ rea = %7。2f\n”程序运行示例如下:I ｎ put r:５。３printf WITHOUT ｗ idth ｏ r precision speci ｆ ications:c ｉｒｃ umferen ｃ e ＝ ３ 3、3 ０ 0854, are ａ = ８８.247 ２６ 3p ｒ in ｔ f WIT Ｈ wi ｄｔｈ ａ nd preci ｓ ion ｓ pecifi ｃａ ti ｏｎ s:ｃ irc ｕ mf ｅ r ｅ nce = 33。3 ０, ａｒｅ a = 88、２ 5＃in ｃｌ ude 〈stdi ｏ。h>ｉ nt main()｛ const do ｕ ble PI=３.14159;ｄｏ uble r; pri ｎ tf(”In ｐ u ｔ r:"); s ｃａ nf(＂%lf”, &ｒ); ｐｒ i ｎ t ｆ("p ｒ intf W Ｉ TH Ｏ UT wiｄ th o ｒ p ｒ ec ｉｓ i ｏ n sp ｅｃ if ｉｃat ｉｏ n ｓ:\n＂);p ｒｉｎ tf(＂ｃｉ rcum ｆ eren ｃ e = ％f, ａ re ａ = ％f\n＂,2＊Ｐ I＊r,Ｐ I＊r*r); print ｆ("ｐ rintf WIT Ｈ ｗｉｄ th and precisio ｎ specifications:\n”);ﻩpr ｉ n ｔ...
