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acm自己做过的一些题

ACM那点事

c++语言中怎么样保留float数小数点的后两位

需要有这个！！！#include

//第一种方式，从根本改变输出流

cout<

//第二种方式，简缩型

cout<

//第三种方式，全局整体型

这种方法是保留两个有效数字,而不是小数点后边两位

cout<

cout.precision(2);

cout<

return 0;

字符与ascii码转换问题

#include

using namespace std;

int main(){

char s='A';

cout<<(int)s<

int x=70;

cout<<(char)x<

}

1006 输入一串字符，判断有多少个重复的字符串

例如：abcd 1

aaaa 4

ababab 3

#include

char s[1000001];

int next[1000001];

int main()

{

int i,j,n,t;

while(gets(s) && s[0]!='.')

{

i=0;

next[0]=-1;

j=-1;

l1:while(s[i]!='\0')

{

if(j==-1 || s[i]==s[j])

{ ++i;

++j;

next[i]=j;

goto l1;

}

j=next[j];

}

n=i;

t=n-next[n];

if(n%t==0)printf("%d\n",n/t);

else printf("1\n");

}

return 0;

}

1009 棋盘问题

... X.. X.O X.O X.O X.O X.O X.O ... ... ... ... .O. .O. OO. OO. ... ... ... ..X ..X X.X X.X XXX

输入：

X.O

OO.

XXX

O.X

XX.

OOO

输出：

yes

#include

using namespace std;

char s[3][4];

int num(char c)

{

int i,j;

int sum=0;

for(i=0;i<3;i++)

for(j=0;j<3;j++)

if(s[i][j]==c)++sum;

return sum;

}

int line(char c)

{

int i;

for(i=0;i<3;i++)

{

if(s[i][0]==s[i][1] && s[i][0]==s[i][2] && s[i][0]==c)return 1;

if(s[0][i]==s[1][i] && s[0][i]==s[2][i] && s[0][i]==c)return 1;

}

if(s[0][0]==s[1][1] && s[0][0]==s[2][2] && s[0][0]==c)return 1;

if(s[0][2]==s[1][1] && s[2][0]==s[1][1] && s[1][1]==c)return 1;

return 0;

}

int main()

{

int a,b;

int n;

cin>>n;

for(;n>0;n--)

{

getchar();

gets(s[0]);

gets(s[1]);

gets(s[2]);

a=num('X');b=num('O');

if(a

{

cout<<"no"<

continue;

}

if(a-b>1)

{

cout<<"no"<

continue;

}

if(a-b==1)

{

if(line('O'))

{

cout<<"no"<

continue;

}

if(a==b)

{

if(line('X'))

{

cout<<"no"<

continue;

}

cout<<"yes"<

}

return 1;

}

1011 矩形小木块，落法，求最小表面积

Sample Input

100

Sample Output

130

#include

using namespace std;

int main()

{

int x;

int area;

int n;

while(cin>>n && n!=0)

{

while(n)

{

cin>>x;

int s=2+4*x;

for(int i=1;i<=x;i++)

{

if(x%i==0)

{

for(int j=1;j<=x/i;j++)

{

if(x/i%j==0)

{

area=2*(i*j+i*x/i/j+j*x/i/j);

if(area

s=area;

}

cout<

n--;

}

1016 一道有关欧拉函数5的问题

Description

It is said that 90% of frosh expect to be above average in their class. You are to provide a reality check.

Input

The first line of standard input contains an integer C, the number of test cases. C data sets follow. Each data set begins with an integer, N, the number of people in the class (1 <= N <= 1000). N integers follow, separated by spaces or newlines, each giving the final grade (an integer between 0 and 100) of a student in the class.

Output

For each case you are to output a line giving the percentage of students whose grade is above average, rounded to 3 decimal places.

Sample Input

5 50 50 70 80 100 7 100 95 90 80 70 60 50 3 70 90 80 3 70 90 81 9 100 99 98 97 9

6 95 94 93 91

Sample Output

40.000%

57.143%

33.333%

66.667%

55.556%

#include

using namespace std;

int main(){

double a[1000],sum,num;;

int t,n,i;

float k;

cin>>t;

for(;t>0;t--)

{

cin>>n;

sum=0;num=0;

for(i=0;i

{

cin>>a[i];

sum+=a[i];

}

sum/=n;

for(i=0;i

if(a[i]>sum)

++num;

k=100*num/n;

cout<

}

return 0;

}

1022 一场战争，持续n天。这个人的上下左右四个方向可以与之发生战争。其中R战胜S，s战胜p，p战胜r。战胜每天一总结，只输出最后的。

Sample Input

3 3 1 RRR

RSR

RRR

3 4 2 RSPR

SPRS

PRSP

Sample Output

RRR

RRRS

RRSP

RSPR

#include

using namespace std;

int main()

{

int l;

cin>>l;

while(l)

{

int x,y,z;

cin>>x>>y>>z;

char a[101][101],b[101][101];

for(int i=0;i

for(int j=0;j

{

cin>>a[i][j];

b[i][j]=a[i][j];

}

while(z)

{

for(int i=0;i

{

for(int j=0;j

{

if(a[i][j]=='R')

{

if(a[i-1][j]=='P' || a[i+1][j]=='P' || a[i][j+1]=='P' || a[i][j-1]=='P')

{

b[i][j]='P';

}

if(a[i][j]=='S')

{

if(a[i-1][j]=='R' || a[i+1][j]=='R' || a[i][j+1]=='R' || a[i][j-1]=='R')

{

b[i][j]='R';

}

if(a[i][j]=='P')

{

if(a[i-1][j]=='S' || a[i+1][j]=='S' || a[i][j+1]=='S' || a[i][j-1]=='S')

{

b[i][j]='S';

}

for(int i=0;i

{

for(int j=0;j

{

a[i][j]=b[i][j];

}

z--;

for(int i=0;i

{

for(int j=0;j

{

cout<

}

cout<

}

if(l>1)

cout<

l--;

}

1027 猜数字，大了输出头too high，小了输出too low，到最后看看一开始顶数字的那个人有没有玩赖。

Sample Input

too high 3

too low 4

too high 2

right on 5

too low 7

too high 6

right on 0

Sample Output

Stan is dishonest

Stan may be honest

#include

using namespace std;

{

int x;

int i;

char bh[30];

string s;

int n[10];

for(i=0;i<10;i++)

{

n[i]=1;

}

while(cin>>x)

{

cin.ignore(100, '\n');

if(x==0) break;

gets(bh);

s=bh;

if(s=="too high")

{

for(i=x-1;i<10;i++)

{

n[i]=0;

}

if(s=="too low")

{

for(i=0;i<=x-1;i++)

{

n[i]=0;

}

if(s=="right on")

{

for(i=0;i<10;i++)

{

if(n[x-1]==1)

{

for(i=0;i<10;i++)

{

n[i]=1;

}

cout<<"Stan may be honest"<

break;

}

else

{

cout<<"Stan is dishonest"<

for(i=0;i<10;i++)

{

n[i]=1;

}

break;

}

1029 锁头转动度数的问题，规则如下：

turn the dial clockwise 2 full turns

stop at the first number of the combination

turn the dial counter-clockwise 1 full turn

continue turning counter-clockwise until the 2nd number is reached

turn the dial clockwise again until the 3rd number is reached

pull the shank and the lock will open.

Sample Input

0 30 0 30 5 35 5 35 0 20 0 20 7 27 7 27 0 10 0 10 9 19 9 19 0 0 0 0

Sample Output

1350

1620

1890

#include

using namespace std;

int main()

{

int a,b,c,d;

int sum;

while(cin>>a>>b>>c>>d && (a || b || c || d))

{

sum=0;

sum=1080+(40+(a-b))%40*9+(40+(c-b))%40*9+(40+(c-d))%40*9;

cout<

}

1029another idea

#include

using namespace std;

float lockemu(int s,int t,int d)

{

int m[40];

for(int i=0;i<=39;i++){m[i]=i;}

int cnt=0,i=s;

while(m[i]!=t)

{

cnt++;

i+=d;

if (i==-1) {i=39;}

if (i==40) {i=0;}

}

return (float)(cnt/40.0)*360;

}

int main()

{

int s1,s2,s3,s4;

int d=0;

while(cin>>s1>>s2>>s3>>s4 && s1+s2+s3+s4!=0)

{

cout<<720+lockemu(s1,s2,-1)+360+lockemu(s2,s3,1)+lockemu(s3,s4,-1)<

}

return 0;

}

1029myself

#include

using namespace std;

int main()

{

int a,b,c,d;

int sum1,sum2,sum3;

int sum;

while(cin>>a>>b>>c>>d)

{

sum1=sum2=sum3=0;

sum=0;

if(a!=0 || b!=0 || c!=0 || d!=0)

{

if(a

sum1=(40-(b-a))*9;

else if(a>b)

sum1=(a-b)*9;

else

sum1=360;

if(b>c)

sum2=(40-(b-c))*9;

else if(b

sum2=(c-b)*9;

else

sum2=360;

if(c

sum3=(40-(d-c))*9;

else if(c>d)

sum3=(c-d)*9;

else

sum3=360;

sum=1080+sum1+sum2+sum3;

cout<

}

else

break;

}

return 0;

}

1039 求m

c的问题

#include

using namespace std;

int main()

{

long long x,y,sum;

while(cin>>x>>y)

{

if(x==0 && y==0)

break;

int i,j;

sum=1;

if(x>y*2)

{

for( i=(x-y+1),j=1;i<=x,j<=y;i++,j++)

{

sum=sum*i/j;

}

if(x<=y*2)

{

for(i=y+1,j=1;i<=x,j<=(x-y);i++,j++)

{

sum=sum*i/j;

}

cout<

}

1159 递推问题，公式如下：F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2)得到的F（n）能被3整除，输出yes，反之输出no。

Sample Input

Sample Output

yes

#include

using namespace std;

int main()

{

int n;

while(cin>>n)

{

if(n<1000000 && n>=0)

{

int a[99999];

a[0]=7;

a[1]=11;

int i;

for(i=2;i<=n;i++)

{

a[i]=a[i-1]+a[i-2];

}

if(i<2)

cout<<"no"<

if(a[i-1]%3==0)

cout<<"yes"<

else

cout<<"no"<

}

else

continue;

}

1060 正明任何一个大于4的整数都可以分解成两个素数相加的和

Sample Input

Sample Output

8 = 3 + 5

20 = 3 + 17

42 = 5 + 37

#include

using namespace std;

int main()

{

int n,m,i,j,k,p;

while(scanf("%d",&n)!=EOF)

{

if(n%2!=0)

continue;

p=n/2;

for(i=2;i<=p;i++)

{

k=0;

for(j=2;j<=(int)sqrt(i);j++)

{

if(i%j==0)

break;

}

if(j==(int)sqrt(i)+1)

{

m=n-i;

for(k=2;k<=(int)sqrt(m);k++)

{

if(m%k==0)

break;

}

if(k==(int)sqrt(m)+1)

{

//cout<

printf("%d = %d + %d\n",n,i,m);

break;

}

if(i==p+1)

printf("Goldbach's conjecture is wrong.");

}

1079 判断字符串包含问题

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia caseDoesMatter CaseDoesMatter Sample Output

Yes

#include

using namespace std;

int main()

{

string s,s1;

while(cin>>s1>>s)

{

int t=0,i=0,j=0;

int length=0,length1=0;

while(s[length])

{

length++;

}

while(s1[length1])

{

length1++;

}

while(s[i])

{

if(s1[j]==s[i])

{

j++;

}

i++;

if(j==length1)

{

t=1;

break;

}

if(t==1)

cout<<"Yes"<

else

cout<<"No"<

}

1082 求两个点（坐标形式）的终点问题

Sample Input

5 10 2 18 2 22

6 14 18 10 18

3 -

4 6 -2 4 -2 6

3 -8 12

4 8 6 12

Sample Output

5 14.000000 2.000000 20.000000 4.000000 18.000000 12.000000 12.000000 18.000000 10.000000 10.000000

3 -3.000000 5.000000 -2.000000 5.000000 -3.000000 6.000000

3 -2.000000 10.000000 5.000000 10.000000 -1.000000 12.000000

#include

using namespace std;

int main()

{

int n;

while(cin>>n && n!=0)

{

int i=0,s;

double a[99999],b[99999];

double c[99999],d[99999];

s=n;

while(s)

{

cin>>a[i]>>b[i];

i++;

s--;

}

for(int j=0;j

{

//cout<<"^^^^^^^^^^^";

if(j

{

c[j]=(a[j]+a[j+1])/2.0;

d[j]=(b[j]+b[j+1])/2.0;

}

if(j==n-1)

{

c[j]=(a[j]+a[0])/2.0;

d[j]=(b[j]+b[0])/2.0;

}

cout<

for(int j=0;j

{

cout<

cout<<" ";

cout<

if(j!=n-1)

cout<<" ";

}

cout<

}

1198 限定一个最小的矩形，把输入的所有的点包含进去

Sample Input

12 56

23 56

13 10

0 0

12 34

0 0

Sample Output

12 10 23 56

12 34 12 34

#include

using namespace std;

int main()

{

int x,y;

int x1,y1,x2,y2;

int i=1,j;

while(i)

{

x1=99999;

y1=99999;

y2=0;

x2=0;

j=0;

while(cin>>x>>y && (x || y))

{

j++;

if(x1>x)

{x1=x;}

if(x2

{x2=x;}

if(y1>y)

{y1=y;}

if(y2

{y2=y;}

}

if(j==0)

break;

cout<

}

1199 一个自定义的加法规则比如：24+1 -> 42+1 -> 43 -> 34

Sample Input

24 1

4358 754

305 794

Sample Output

1998

#include