1.加法器和累乘器1#include<stdio.h>23#defineMAXN11045intget_sum(int*arr,intn)6{7intsum=0;//初值為08for(inti=0;i<n;i++)9sum+=*(arr+i);10returnsum;11}1213intget_mul(int*arr,intn)14{15intmul=1;//初值為116for(inti=0;i<n;i++)17mul*=*(arr+i);18returnmul;19}2021intmain()22{23intn,a[MAXN];24scanf("%d",&n);25for(inti=0;i<n;i++)26scanf("%d",&a[i]);2728intres1=get_sum(a,n),res2=get_mul(a,n);2930printf("%d%d\n",res1,res2);31return0;32}2.求最大數與最小數1#include<stdio.h>23#defineMAXN11045intmax(inta,intb)//求兩數最大值6{7returna>b?a:b;8}910intmin(inta,intb)//求兩數最小值11{12returna<b?a:b;13}1415intget_max(int*arr,intn)//求數組的最大值16{17intmaxx=*arr;18for(inti=1;i<n;i++)19if(*(arr+i)>maxx)20maxx=*(arr+i);21returnmaxx;22}2324intget_min(int*arr,intn)//求數組的最小值25{26intminx=*arr;27for(inti=1;i<n;i++)28if(*(arr+i)<minx)29minx=*(arr+i);30returnminx;31}No.1/23323334intmain()35{36intn,a[MAXN];37scanf("%d",&n);38for(inti=0;i<n;i++)39scanf("%d",&a[i]);4041intres1=get_max(a,n),res2=get_min(a,n);4243printf("%d%d\n",res1,res2);4445intx,y;46scanf("%d%d",&x,&y);47intres3=max(x,y),res4=min(x,y);48printf("%d%d\n",res3,res4);49return0;50}3.排序（冒泡排序、選擇排序等）1#include<stdio.h>23#defineMAXN11045voidbuble_sort(int*arr,intn)//冒泡排序(升序)6{7inttemp;8for(inti=0;i<n;i++)9for(intj=0;j<n-1-i;j++)//注意j的范圍10if(*(arr+j)>*(arr+j+1))//降序改為<11{12temp=*(arr+j);13*(arr+j)=*(arr+j+1);14*(arr+j+1)=temp;15}16}1718intmain()19{20intn,a[MAXN];21scanf("%d",&n);22for(inti=0;i<n;i++)23scanf("%d",&a[i]);2425buble_sort(a,n);2627for(inti=0;i<n;i++)28printf("%d",a[i]);2930return0;31}1#include<stdio.h>23#defineMAXN11045voidselect_sort(int*arr,intn)//冒泡排序(升序)6{7inttemp;8for(inti=0;i<n;i++)9{10intpos=i;//找到當前的最小(大)值11for(intj=i+1;j<n;j++)No.2/2312if(*(arr+j)<*(arr+pos))//降序改為<13pos=j;14temp=*(arr+i);15*(arr+i)=*(arr+pos);16*(arr+pos)=temp;17}18}1920intmain()21{22intn,a[MAXN];23scanf("%d",&n);24for(inti=0;i<n;i++)25scanf("%d",&a[i]);2627select_sort(a,n);2829for(inti=0;i<n;i++)30printf("%d",a[i]);3132return0;33}4.大小寫字母轉換1#include<stdio.h>23#defineMAXN11045charget_c(charc)6{7if(c>='A'&&c<='Z')//大寫轉小寫8returnc+32;9elseif(c>='a'&&c<='z')//小寫轉大寫10returnc-32;11returnc;//其他字符照常返回12}1314voidget_str(char*str)15{16inti=0;17while(*(str+i)!='\0')18{19*(str+i)=get_c(*(str+i));20i++;21}22}2324intmain()25{26chars[MAXN];27gets(s);28get_str(s);29puts(s);30return0;31}5.判別鍵盤輸入字符的類別1#include<stdio.h>23#defineMAXN11045intcount_num=0,count_letter=0,count_space=0,count_other=0;6//開全局,懶得在函數中設置指針傳值78voidget_count(char*str)9{10inti=0;No.3/2311while(*(str+i)!='\0')12{13charc=*(str+i);14if(c>='0'&&c<='9')//數字15count_num++;16elseif(c=='')//空格17count_space++;18elseif((c>='A'&&c<='Z')||(c>='a'&&c<='z'))//字母19//elseif(c>='A'&&c<='Z'||c>='a'&&c<='z')20//此處由于&&優先級高于||可不用加括號21count_letter++;22elsecount_other++;//其他23i++;24}25}2627intmain()28{29chars[MAXN];30gets(s);31get_count(s);32printf("%d%d%d%d\n",count_num,count_letter,count_space,count_other);3334return0;35}6.判別閏年1#include<stdio.h>23intis_leap(intyear)4{5if((year%4==0&&year%100!=0)||year%400==0)6return1;7return0;8}910intmain()11{12intyear;13scanf("%d",&year);1415printf(is_leap(year)?"Yes":"No");1617return0;18}7.百分制成績與等級制成績互相轉換1#include<stdio.h>23intmain()4{5intscore;6scanf("%d",&score);7switch(score/10)8{9case10://此處注意考慮100分10case9:printf("A");break;11case8:printf("B");break;12case7:printf("C");break;13case6:printf("D");break;14default:printf("E");15}16//if(score>=90)17//printf("A");18//elseif(score>=80)19//printf("B");20//elseif(score>=70)No.4/2321//printf("C");22//elseif(score>=60)23//printf("D");24//elseprintf("E");25return0;26}8.求兩個數的最大公因數和最小公倍數1#include<stdio.h>23intgcd(inta,intb)4{5returnb==0?a:gcd(b,a%b);6}78intlcm(inta,intb)9{10returna/gcd(a,b)*b;//先除后乘,防止乘爆11}1213intmain()14{15intx,y;16scanf("%d%d",&x,&y);17printf("%d%d\n",gcd(x,y),lcm(x,y));18return0;19}9.求菲比拉契數列有限項1#include<stdio.h>23intf1(intn)//遞歸寫法4{5if(n<=2)6return1;7returnf1(n-1)+f1(n-2);8}910intf2(intn)//循環(非數組)11{12if(n<=2)13return1;14inta=1,b=1,c;15for(inti=3;i<=n;i++)16c=a+b,a=b,b=c;17returnb;18}1920#defineMAXN11021intarr[MAXN];2223intf3(intn)//循環(數組)24{25arr[1]=arr[2]=1;26for(inti=3;i<=n;i++)27arr[i]=arr[i-1]+arr[i-2];28returnarr[n];29}3031intmain()32{33intn;34scanf("%d",&n);3536printf("%d\n",f1(n));3738printf("%d\n",f2(n));No.5/233940printf("%d\n",f3(n));4142return0;43}10.統計學生成績，包括總成績、平均成績、各分數段人數等1#include<stdio.h>23#defineMAXN11045intcount[11];//統計0~100分數(每10分一段)67intget_sum(int*arr,intn)8{9intres=0;10for(inti=0;i<n;i++)11res+=*(arr+i);12returnres;13}1415voidget_count(int*arr,intn)16{17for(inti=0;i<n;i++)18count[*(arr+i)/10]++;19}2021intmain()22{23intn;24inta[MAXN];25scanf("%d",&n);26for(inti=0;i<n;i++)27scanf("%d",&a[i]);2829intsum=get_sum(a,n);3031doubleave=sum*1./n;//轉為double,可通過*1.強制類型轉換3233printf("%d%lf\n",sum,ave);34get_count(a,n);3536for(inti=0;i<=10;i++)37printf("%d",count[i]);3839return0;40}11.驗證哥德巴赫猜想1例如：28=3+5320=3+17=7+13442=5+37=11+31=13+29=19+231#include<stdio.h>2#include<cmath>34intis_prime(intx)5{6if(x==1)7return0;8for(inti=2;i<sqrt(x);i++)9if(x%i==0)10return0;No.6/2311return1;12}1314intmain()15{16intn;17scanf("%d",&n);18for(inti=2;i<n;i++)19if(is_prime(i)&&is_prime(n-i))20{21printf("%d=%d+%d\n",n,i,n-i);22//break;23}24return0;25}12.用窮舉法求某數段的素數、水仙花數、完全平方數等。1#include<stdio.h>2#include<math.h>34intis_prime(intx)//判斷素數5{6if(x==1)7return0;8for(inti=2;i<sqrt(x);i++)9if(x%i==0)10return0;11return1;12}1314voidget_primes(intl,intr)//求素數15{16for(inti=l;i<=r;i++)17if(is_prime(i))18printf("%d",i);19printf("\n");20}2122voidget_res2(intl,intr)//水仙花數23{24for(inti=l;i<=r;i++)25{26inta=i/100,b=i/10%10,c=i%10;27if(a*a*a+b*b*b+c*c*c==i)28printf("%d",i);29}30printf("\n");31}3233voidget_res3(intl,intr)//求平方數34{35for(inti=l;i<=r;i++)36{37intsq=sqrt(i);38if(sq*sq==i)39printf("%d",i);40}41printf("\n");42}434445intmain()46{47inta,b;48scanf("%d%d",&a,&b);4950get_primes(a,b);No.7/2351get_res2(a,b);5253get_res3(a,b);5455return0;56}13.求近似數（如定積分、用牛頓迭代法或二分法或弦截法求多元方程的根）:,,,)：1#include<stdio.h>2#include<math.h>34doublef(doublex)5{6returnsin(x)*cos(x);7}89intmain()10{11doublel,r;12intn;13scanf("%lf%lf",&l,&r);14scanf("%d",&n);1516doubleh=(r-l)/n;17doubles=h/2*(f(l)+f(r));18for(inti=1;i<=n-1;i++)19s+=f(l+i*h)*h;2021printf("%.4f\n",s);22return0;23}過

把過根.No.8/23：令.:.1#include<stdio.h>2#include<math.h>34constdoubleeps=1e-5;//誤差5//f=2x^3-4x^2+3x-66doublef(doublex)7{8return2*x*x*x-4*x*x+3*x-6;9}10//f'=6x^2-8x+311doublef_(doublex)12{13return6*x*x-8*x+3;14}1516intmain()17{18doublex;19scanf("%lf",&x);20doublex_0=x,x_1=x_0-f(x_0)/f_(x_0);21while(fabs(x_1-x_0)>=eps)22{23x_0=x_1;24x_1=x_0-f(x_0)/f_(x_0);25}2627printf("%.4f\n",x_1);28return0;29}和在

若和則在和在和.:.1#include<stdio.h>2#include<math.h>34constdoubleeps=1e-5;//誤差5//f=2x^3-4x^2+3x-66doublef(doublex)7{8return2*x*x*x-4*x*x+3*x-6;9}1011intmain()12{13doublel,r,mid;14scanf("%lf%lf",&l,&r);1516while(l+eps<r)17{18mid=(l+r)/2;19//if(fabs(f(mid))<eps)20//break;21if(f(l)*f(mid)<=0)//更新右端點22r=mid;23elsel=mid;24}25No.9/2326printf("%.4f\n",l);//此處r亦可27return0;28}設令.:.1#include<stdio.h>2#include<math.h>34constdoubleeps=1e-5;//誤差5//f=2x^3-4x^2+3x-66doublef(doublex)7{8return2*x*x*x-4*x*x+3*x-6;9}1011intmain()12{13doublel,r;14scanf("%lf%lf",&l,&r);1516doublex=(l*f(r)-r*f(l))/(f(r)-f(l));17while(fabs(f(x))>eps)18{19l=r;20r=x;21x=(l*f(r)-r*f(l))/(f(r)-f(l));22}2324printf("%.4f\n",x);25return0;No.10/2326}14.求兩個矩陣之和、之積1#include<stdio.h>23#defineMAXN11045intn,m;67voidget_matrix(inta[][MAXN],intb[][MAXN],intc[][MAXN])8{9for(inti=0;i<n;i++)10for(intj=0;j<m;j++)11c[i][j]=a[i][j]+b[i][j];12}1314intmain()15{1617inta[MAXN][MAXN],b[MAXN][MAXN],sum[MAXN][MAXN];1819scanf("%d%d",&n,&m);//矩陣同型才能求和2021for(inti=0;i<n;i++)22for(intj=0;j<m;j++)23scanf("%d",&a[i][j]);24for(inti=0;i<n;i++)25for(intj=0;j<m;j++)26scanf("%d",&b[i][j]);2728//for(inti=0;i<n;i++)29//for(intj=0;j<m;j++)30//sum[i][j]=a[i][j]+b[i][j];31get_matrix(a,b,sum);3233for(inti=0;i<n;i++)34{35for(intj=0;j<m;j++)36printf("%d",sum[i][j]);37printf("\n");38}39return0;40}1#include<stdio.h>23#defineMAXN11045intn,m,p;67voidget_matrix(inta[][MAXN],intb[][MAXN],intc[][MAXN])8{9for(inti=0;i<n;i++)10for(intj=0;j<p;j++)11{12c[i][j]=0;13for(intk=0;k<m;k++)14c[i][j]+=a[i][k]*b[k][j];15}16}1718intmain()19{20No./2321inta[MAXN][MAXN],b[MAXN][MAXN],mul[MAXN][MAXN];2223scanf("%d%d%d",&n,&m,&p);//矩陣n*mm*p求積2425for(inti=0;i<n;i++)26for(intj=0;j<m;j++)27scanf("%d",&a[i][j]);28for(inti=0;i<m;i++)29for(intj=0;j<p;j++)30scanf("%d",&b[i][j]);3132//for(inti=0;i<n;i++)33//for(intj=0;j<p;j++)34//{35//mul[i][j]=0;36//for(intk=0;k<m;k++)37//mul[i][j]+=a[i][k]*b[k][j];38//}39get_matrix(a,b,mul);404142for(inti=0;i<n;i++)43{44for(intj=0;j<p;j++)45printf("%d",mul[i][j]);46printf("\n");47}48return0;49}15.統計輸入字符中的單詞個數1#include<stdio.h>2#include<string.h>34#defineMAXN11056intmain()7{8chars[MAXN];9intres=0;10gets(s);1112intlen=strlen(s);13for(inti=0;i<len;i++)14{15if(s[i]!='')//不是空格16{17intj=i;18while(j<len&&s[j]!='')19j++;20res++;21i=j;22}23}2425printf("%d\n",res);2627return0;28}16.最近點對問題(分治)(1///problem/P14292No.12/233#include<stdio.h>4#include<cmath>5#defineMAXN20001067constdoubleinf=0x3f3f3f3f,eps=1e-5;89structPoint{10doublex;11doubley;12}p[MAXN];1314doubleget_dist(structPointa,structPointb)15{16doubledx=a.x-b.x,dy=a.y-b.y;17returnsqrt(dx*dx+dy*dy);18}1920intmain()21{22intn;23doubleres=inf;24scanf("%d",&n);25for(inti=0;i<n;i++)26scanf("%lf%lf",&p[i].x,&p[i].y);2728for(inti=0;i<n;i++)//枚舉兩個點29for(intj=i+1;j<n;j++)30{31doubledist=get_dist(p[i],p[j]);32if(dist+eps<res)//精度保證33res=dist;34}3536printf("%.4f\n",res);37return0;38}()將(,)=令即故.,(.在,

:),,No.13/23,:將,很

.c++.1/////problem/P14292#include<iostream>3#include<cstring>4#include<algorithm>5#include<cmath>67#defineMAXN20001089usingnamespacestd;1011constdoubleinf=0x3f3f3f3f,eps=1e-5;1213intn;14doubleres=inf;15structPoint{16doublex;17doubley;No.14/2318}p[MAXN],temp1[MAXN],temp2[MAXN];1920intcmp_1(Pointa,Pointb)//C++struct在非定義時可不寫21{22if(a.x==b.x)//按照y排序23returna.y<b.y;24returna.x<b.x;///按照x排序25}2627intcmp_2(Pointa,Pointb)28{29returna.y<b.y;30}3132doubleget_dist(Pointa,Pointb)33{34doubledx=a.x-b.x,dy=a.y-b.y;35returnsqrt(dx*dx+dy*dy);36}373839voidmerge(intl,intmid,intr)//歸并排序40{41intcnt=0;42inti=l,j=mid+1;43while(i<=mid&&j<=r)44{45if(p[i].y<p[j].y)46temp1[cnt++]=p[i],i++;47elsetemp1[cnt++]=p[j],j++;48}49while(i<=mid)50temp1[cnt++]=p[i],i++;51while(j<=r)52temp1[cnt++]=p[j],j++;53for(inti=0;i<cnt;i++)54p[l+i]=temp1[i];55}5657voiddfs(intl,intr)58{59//cout<<l<<""<<r<<endl;60if(r-l<=3)61{62for(inti=l;i<=r;i++)63for(intj=i+1;j<=r;j++)64{65doubledist=get_dist(p[i],p[j]);66if(dist+eps<res)67res=dist;68}69sort(p+l,p+r+1,cmp_2);//根據y排序70return;71}7273intmid=(l+r)/2;7475767778intmid_x=p[mid].x;dfs(l,mid),dfs(mid+1,r);//遞歸左右兩邊merge(l,mid,r);intcnt=0;798081828384858687for(inti=l;i<=r;i++){if(fabs(p[i].x-mid_x)+eps<res){//倒著枚舉for(intj=cnt-1;j>=0&&p[i].y-temp2[j].y<res;j--){doubledist=get_dist(p[i],temp2[j]);if(dist+eps<res)No.15/2388res=dist;89}90temp2[cnt++]=p[i];91}92}93}9495intmain()96{97scanf("%d",&n);98for(inti=0;i<n;i++)99scanf("%lf%lf",&p[i].x,&p[i].y);100101sort(p,p+n,cmp_1);102dfs(0,n-1);103104printf("%.4f\n",res);105106return0;107}17.背包問題(動態規劃)有第.前)有.(:...1for(inti=1;i<=n;i++)2for(intj=0;j<=V;j++)3{4if(j<v[i])5dp[i][j]=dp[i-1][j];6elsedp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);7}.與.:.與是.:.No.16/231for(inti=1;i<=n;i++)2for(intj=V;j>=0;j--){3if(j<v[i])4dp[j]=dp[j];5elsedp[j]=max(dp[j],dp[j-v[i]]+w[i]);6}.1for(inti=1;i<=n;i++)2for(intj=V;j>=v[i];j--)3dp[j]=max(dp[j],dp[j-v[i]]+w[i]);.1for(inti=1;i<=n;i++){2intv,w;3scanf("%d%d",&v,&w);4for(intj=V;j>=v;j--)5dp[j]=max(dp[j],dp[j-v]+w);6}有第.同前第..1for(inti=1;i<=n;i++)2for(intj=0;j<=V;j++){3intamount=j/v[i];4for(intk=0;k<=amount;k++)5dp[i][j]=max(dp[i][j],dp[i-1][j-k*v[i]]+k*w[i]);6}.:前.將:..No.17/231for(inti=1;i<=n;i++)2for(intj=0;j<=V;j++){3dp[i][j]=dp[i-1][j];4if(j>=v[i])5dp[i][j]=max(dp[i][j],dp[i][j-v[i]]+w[i]);6//dp[i][j]=max(dp[i-1][j],dp[i][j-v[i]]+w[i]);7}1for(inti=1;i<=n;i++)2for(intj=v[i];j<=V;j++)3dp[j]=max(dp[j],dp[j-v[i]]+w[i]);同1for(inti=1;i<=n;i++){2intv,w;3scanf("%d%d",&v,&w);4for(intj=v;j<=V;j++)5dp[j]=max(dp[j],dp[j-v]+w);6}有第..1for(inti=1;i<=n;i++)2for(intj=0;j<=V;j++)3for(intk=0;k<=s[i]&&k<=j/v[i];k++)4dp[i][j]=max(dp[i][j],dp[i-1][j-v[i]*k]+w[i]*k);個和.,.,

.:No.18/23將為.：..,,,

故

..1#include<iostream>2#include<cstring>3#include<algorithm>45#defineMAXN201067usingnamespacestd;89intn,V,cnt;10intv[10*MAXN],w[10*MAXN];11intdp[MAXN];1213intmain()14{15scanf("%d%d",&n,&V);16for(inti=1;i<=n;i++)17{18inttemp_v,temp_w,s;19scanf("%d%d%d",&temp_v,&temp_w,&s);20for(intk=1;k<=s;k*=2)21{22s-=k;23v[++cnt]=k*temp_v;24w[cnt]=k*temp_w;25}26if(s>0)27v[++cnt]=s*temp_v,w[cnt]=s*temp_w;28}2930for(inti=1;i<=cnt;i++)31for(intj=V;j>=v[i];j--)32dp[j]=max(dp[j],dp[j-v[i]]+w[i]);33printf("%d\n",dp[V]);34return0;35}3618.多機調度問題(貪心)由的.No.19/23當.c++.1#include<iostream>2#include<cstring>3#include<algorithm>4#include<queue>56#defineMAXN1001078usingnamespacestd;910intn,m;11structNode{12inttime;13intid;14booloperator<(constNode&a)const//構造小根堆15{16if(time==a.time)17returnid>a.id;18elsereturntime>a.time