学生成绩得分排行
假設(shè)有一教師依學(xué)生座號(hào)輸入考試分?jǐn)?shù),現(xiàn)希望在輸入完畢后自動(dòng)顯示學(xué)生分?jǐn)?shù)的排行,當(dāng)然學(xué)生的分?jǐn)?shù)可能相同。
這個(gè)問(wèn)題基本上要解不難,只要使用額外的一個(gè)排行陣列走訪分?jǐn)?shù)陣列就可以了,直接使用下面的程式片段作說(shuō)明:
? for (i? = ? 0 ;?i? < ?count;?i ++ )? ... {?
????juni[i]?=?1;?
????for(j?=?0;?j?<?count;?j++)?...{?
????????if(score[j]?>?score[i])?
????????????juni[i]++;?
????}?
} ?
printf( " 得分 排行 " );?
for (i? = ? 0 ;?i? < ?count;?i ++ )?
????printf( " %d %d " ,?score[i],?juni[i]);?
public ? class ?ScoreRank?
... {????
????public?static?void?main(String[]?args)?throws?NumberFormatException,?IOException?
????...{????????
????????final?int?MAX?=?100;????????
????????final?int?MIN?=?0;????????????????
????????int[]?score?=?new?int[MAX+1];?????????
????????int[]?juni?=?new?int[MAX+2];????????????????
????????BufferedReader?reader?=?????????????
????????????new?BufferedReader(new?InputStreamReader(System.in));????????
????????int?count?=?0;?????????
????????do?
????????...{?????????????
????????????System.out.print("輸入分?jǐn)?shù),-1結(jié)束:");?????????????
????????????score[count++]?=?Integer.parseInt(reader.readLine());?????????
????????}?while((score[count-1]?!=?-1)?and?(?(count?-1)?<?MAX?)?);????????
????????count--;?????????
????????for(int?i?=?0;?i?<?count;?i++)?????????????
????????????juni[score[i]]++;?????????
????????juni[MAX+1]?=?1;?????????
????????for(int?i?=?MAX;?i?>=?MIN;?i--)?????????????
????????????juni[i]?=?juni[i]?+?juni[i+1];?????????
????????System.out.println("得分 排行");?????????
????????for(int?i?=?0;?i?<?count;?i++)?
????????...{????????????
????????????System.out.println(score[i]?+?" "?+?juni[score[i]+1]);?????????
????????}????
????}
} ?
?
這個(gè)問(wèn)題基本上要解不難,只要使用額外的一個(gè)排行陣列走訪分?jǐn)?shù)陣列就可以了,直接使用下面的程式片段作說(shuō)明:
? for (i? = ? 0 ;?i? < ?count;?i ++ )? ... {?
????juni[i]?=?1;?
????for(j?=?0;?j?<?count;?j++)?...{?
????????if(score[j]?>?score[i])?
????????????juni[i]++;?
????}?
} ?
printf( " 得分 排行 " );?
for (i? = ? 0 ;?i? < ?count;?i ++ )?
????printf( " %d %d " ,?score[i],?juni[i]);?
上面這個(gè)方法雖然簡(jiǎn)單,但是反覆計(jì)算的次數(shù)是n^2,如果n值變大,那么運(yùn)算的時(shí)間就會(huì)拖長(zhǎng);改變juni陣列的長(zhǎng)度為n+2,并將初始值設(shè)定為0,如下所示:
接下來(lái)走訪分?jǐn)?shù)陣列,并在分?jǐn)?shù)所對(duì)應(yīng)的排行陣列索引元素上加1,如下所示:
將排行陣列最右邊的元素設(shè)定為1,然后依序?qū)⒂疫叺脑刂导又磷筮呉粋€(gè)元素,最后排行陣列中的“分?jǐn)?shù)+1””就是得該分?jǐn)?shù)的排行,如下所示:
這樣的方式看起來(lái)復(fù)雜,其實(shí)不過(guò)在計(jì)算某分?jǐn)?shù)之前排行的人數(shù),假設(shè)89分之前的排行人數(shù)為x人,則89分自然就是x+1了,這也是為什么排行陣列最右邊要設(shè)定為1的原因;如果89分有y人,則88分自然就是x+y+1,整個(gè)陣列右邊元素向左加的原因正是如此。
如果分?jǐn)?shù)有負(fù)分的情況,由于C/C++或Java等程式語(yǔ)言無(wú)法處理負(fù)的索引,所以必須加上一個(gè)偏移值,將所有的分?jǐn)?shù)先往右偏移一個(gè)范圍即可,最后顯示的時(shí)候記得減回偏移值就可以了。
public ? class ?ScoreRank?
... {????
????public?static?void?main(String[]?args)?throws?NumberFormatException,?IOException?
????...{????????
????????final?int?MAX?=?100;????????
????????final?int?MIN?=?0;????????????????
????????int[]?score?=?new?int[MAX+1];?????????
????????int[]?juni?=?new?int[MAX+2];????????????????
????????BufferedReader?reader?=?????????????
????????????new?BufferedReader(new?InputStreamReader(System.in));????????
????????int?count?=?0;?????????
????????do?
????????...{?????????????
????????????System.out.print("輸入分?jǐn)?shù),-1結(jié)束:");?????????????
????????????score[count++]?=?Integer.parseInt(reader.readLine());?????????
????????}?while((score[count-1]?!=?-1)?and?(?(count?-1)?<?MAX?)?);????????
????????count--;?????????
????????for(int?i?=?0;?i?<?count;?i++)?????????????
????????????juni[score[i]]++;?????????
????????juni[MAX+1]?=?1;?????????
????????for(int?i?=?MAX;?i?>=?MIN;?i--)?????????????
????????????juni[i]?=?juni[i]?+?juni[i+1];?????????
????????System.out.println("得分 排行");?????????
????????for(int?i?=?0;?i?<?count;?i++)?
????????...{????????????
????????????System.out.println(score[i]?+?" "?+?juni[score[i]+1]);?????????
????????}????
????}
} ?
?
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