hdu2090
模拟
1 #include<stdio.h>
2 int main(){
3 char b[100];
4 double price=0,a1,a2;
5 int i=0;
6 while(scanf("%s%lf%lf",b,&a1,&a2)!=EOF){
7 price+=a1*a2;
8 // printf("%.1lf
",price);
9 }
10 printf("%.1lf
",price);
11 return 0;
12 }
hdu2091
模拟实现
1 #include<stdio.h>
2
3 int main(){
4 char a;
5 int b;
6 int t=0;
7 while(scanf("%c",&a)!=EOF&&a!='@'){
8 scanf("%d%*c",&b);
9 if(t++)printf("
");
10 int i;
11 if(b==1){
12 printf("%c
",a);
13 continue;
14 }
15 for(i=1;i<=b-1;i++)printf(" ");
16 printf("%c
",a);
17 if(b>=3){
18 for(i=2;i<=b-1;i++){
19 for(int j=1;j<=b-i;j++) printf(" ");
20 printf ("%c",a);
21 for(int j=1;j<=2*(i-1)-1;j++) printf(" ");
22 printf ("%c
",a);
23 }
24 }
25 for(i=1;i<=2*b-1;i++)printf("%c",a);
26 printf("
");
27 }
28 return 0;
29 }
hdu2092
暴力模拟
1 #include<stdio.h>
2 #include<math.h>
3
4 int main(){
5 int m,n;
6 while(scanf("%d%d",&n,&m)!=EOF&&(n!=0||m!=0)){
7 double d=n*n-4*m;
8 if(d<0){
9 printf("No
");
10 continue;
11 }
12 else{
13 double l1=n/2.0+sqrt(d)/2.0,l2=n/2.0-sqrt(d)/2.0;
14 int t1=l1,t2=l2;
15 if(fabs(t1-l1)<1e-10&&fabs(t2-l2)<1e-10)printf("Yes
");
16 else printf("No
");
17 }
18 }
19 return 0;
20 }
hdu2093
麻烦一点的模拟
1 #include<stdio.h>
2 #include<string.h>
3 struct list{
4 char name[10];
5 int am;
6 int pt;
7 }l[7];
8 void ex(int i,int j){
9 char temp[10];int t;
10 {
11 strcpy(temp,l[i].name);
12 strcpy(l[i].name,l[j].name);
13 strcpy(l[j].name,temp);
14 }
15 {
16 t=l[i].am;
17 l[i].am=l[j].am;
18 l[j].am=t;
19 }
20 {
21 t=l[i].pt;
22 l[i].pt=l[j].pt;
23 l[j].pt=t;
24 }
25 }
26 int main(){
27 int p[7];
28 int n,m;
29 while(scanf("%d%d",&n,&m)!=EOF){
30 for(int q=1;q<=6;q++){
31 l[q].am=l[q].pt=0;
32 scanf("%s",l[q].name);
33 for(int k=1;k<=n;k++){
34 int a;
35 scanf("%d",&a);
36 if(a>0){
37 l[q].am++;
38 l[q].pt+=a;
39 }
40 char b=getchar();
41 if(b=='('){
42 scanf("%d",&a);
43 l[q].pt+=m*a;
44 getchar();
45 }
46 }
47
48 }
49 int i,j;
50 for(i=1;i<=5;i++){
51 for(j=i+1;j<=6;j++){
52 if(l[i].am<l[j].am){
53 ex(i,j);
54 }
55 else if(l[i].am==l[j].am&&l[i].pt>l[j].pt){
56 ex(i,j);
57 }
58 else if(l[i].am==l[j].am&&l[i].pt==l[j].pt&&strcmp(l[i].name,l[j].name)>0){
59 ex(i,j);
60 }
61 }
62 }
63 for(i=1;i<=6;i++){
64 printf("%-10s %2d %4d
",l[i].name,l[i].am,l[i].pt);
65 }
66 }
67 return 0;
68 }
hdu2094
有胜负关系，判断是否能够决出冠军（唯一不败），map记录某个人是否失败过
1 #include<stdio.h>
2 #include<string.h>
3 #include<map>
4 #include<string>
5 #include<iostream>
6 using namespace std;
7
8 int main(){
9 int n;
10 while(scanf("%d",&n)!=EOF&&n){
11 map<string,int>m;
12 int ans=0;
13 while(n--){
14 string a,b;
15 cin>>a>>b;
16 if(m[a]==0){
17 ans++;
18 m[a]=-1;
19 }
20 if(m[b]==0)m[b]++;
21 else if(m[b]==-1){
22 m[b]=1;
23 ans--;
24 }
25 }
26 if(ans==1)printf("Yes
");
27 else printf("No
");
28 }
29 return 0;
30 }
hdu2095
找唯一一个只出现一次的数，将所有数异或，最后得到的就是出现一次的数
1 #include<stdio.h>
2 int main(){
3 int n;
4 while(scanf("%d",&n)!=EOF&&n!=0){
5 int a=0,b;
6 for(int q=1;q<=n;q++){
7 scanf("%d",&b);
8 a^=b;
9 }
10 printf("%d
",a);
11 }
12 return 0;
13 }
hdu2096
最后两位的A+B，模拟
1 #include<stdio.h>
2 int main(){
3 int T;
4 while(scanf("%d",&T)!=EOF){
5 for(int q=1;q<=T;q++){
6 int a,b;
7 scanf("%d%d",&a,&b);
8 a%=100;
9 b%=100;
10 int ans=(a+b)%100;
11 printf("%d
",ans);
12 }
13
14 }
15 return 0;
16 }
hdu2097
模拟
1 #include<stdio.h>
2
3 int main(){
4 int n;
5 while(scanf("%d",&n)!=EOF&&n!=0){
6 int ans10=0,ans12=0,ans16=0;
7 int t=n;
8 while(t){
9 ans10+=t%10;
10 t/=10;
11 }
12 t=n;
13 while(t){
14 ans16+=t%16;
15 t/=16;
16 }
17 t=n;
18 while(t){
19 ans12+=t%12;
20 t/=12;
21 }
22 if(ans10==ans12&&ans12==ans16)printf("%d is a Sky Number.
",n);
23 else printf("%d is not a Sky Number.
",n);
24 }
25 return 0;
26 }