Recently, paleoanthropologists have found historical remains on an island in the Atlantic Ocean. The most inspiring thing is that they excavated in a magnificent cave and found that it was a huge tomb. Inside the construction,researchers identified a large number of skeletons, and funeral objects including stone axe, livestock bones and murals. Now, all items have been sorted, and they can be divided into N types. After they were checked attentively, you are told that there areAi items of the i-th type. Further more, each item of the i-th type requires Bi million dollars for transportation, analysis, and preservation averagely. As your job, you need to calculate the total expenditure.
The first line of input contains an integer T which is the number of test cases. For each test case, the first line contains an integer N which is the number of types. In the next N lines, the i-th line contains two numbers Ai and Bi as described above. All numbers are positive integers and less than 101.
For each case, output one integer, the total expenditure in million dollars.
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
int t;
int a,b;
scanf("%d",&t);
for(int i=0;i<t;i++){
scanf("%d",&n);
int sum=0;
for(int j=0;j<n;j++){
scanf("%d %d",&a,&b);
sum+=a*b;
}
printf("%d\n",sum);
}
return 0;
}
B、Pocket Cube_5983
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 23 Accepted Submission(s): 9
Problem Description
The Pocket Cube, also known as the Mini Cube or the Ice Cube, is the 2 × 2 × 2 equivalence of a Rubik’s Cube.
The cube consists of 8 pieces, all corners.
Each piece is labeled by a three dimensional coordinate (h, k, l) where h, k, l ∈ {0, 1}. Each of the six faces owns four small faces filled with a positive integer.
For each step, you can choose a certain face and turn the face ninety degrees clockwise or counterclockwise.
You should judge that if one can restore the pocket cube in one step. We say a pocket cube has been restored if each face owns four same integers.
Input
The first line of input contains one integer N(N ≤ 30) which is the number of test cases.
For each test case, the first line describes the top face of the pocket cube, which is the common 2 × 2 face of pieces
labelled by (0, 0, 1),(0, 1, 1),(1, 0, 1),(1, 1, 1). Four integers are given corresponding to the above pieces.
The second line describes the front face, the common face of (1, 0, 1),(1, 1, 1),(1, 0, 0),(1, 1, 0). Four integers are
given corresponding to the above pieces.
The third line describes the bottom face, the common face of (1, 0, 0),(1, 1, 0),(0, 0, 0),(0, 1, 0). Four integers are
given corresponding to the above pieces.
The fourth line describes the back face, the common face of (0, 0, 0),(0, 1, 0),(0, 0, 1),(0, 1, 1). Four integers are
given corresponding to the above pieces.
The fifth line describes the left face, the common face of (0, 0, 0),(0, 0, 1),(1, 0, 0),(1, 0, 1). Four integers are given
corresponding to the above pieces.
The six line describes the right face, the common face of (0, 1, 1),(0, 1, 0),(1, 1, 1),(1, 1, 0). Four integers are given
corresponding to the above pieces.
In other words, each test case contains 24 integers a, b, c to x. You can flat the surface to get the surface development
as follows.
+ - + - + - + - + - + - +
| q | r | a | b | u | v |
+ - + - + - + - + - + - +
| s | t | c | d | w | x |
+ - + - + - + - + - + - +
| e | f |
+ - + - +
| g | h |
+ - + - +
| i | j |
+ - + - +
| k | l |
+ - + - +
| m | n |
+ - + - +
| o | p |
+ - + - +
Output
For each test case, output YES if can be restored in one step, otherwise output NO.
Sample Input
4 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 6 6 1 1 1 1 2 2 2 2 3 3 3 3 5 5 5 5 4 4 4 4 1 4 1 4 2 1 2 1 3 2 3 2 4 3 4 3 5 5 5 5 6 6 6 6 1 3 1 3 2 4 2 4 3 1 3 1 4 2 4 2 5 5 5 5 6 6 6 6
Sample Output
Source
Recommend
jiangzijing2015 | We have carefully selected several similar problems for you:
5994 5993 5992 5991 5990 思路：在比赛的时候理解错意思了，以为是判断能不能复原模仿，而没看到只需要一步。
魔方是2*2的，有三种转的方法，上面顺时针转90度，逆时针转，前面顺时针，前面逆时针，左边顺时针（相当于右边顺时针），左边逆时针（相当于左边逆时针）
还有再加上不操作算是一种，总共七种类，模拟出来就行了。
1 #include <iostream>
2 #include <cstdio>
3
4 using namespace std;
5
6 int a[7][5];
7 int b[7][5];
8
9 bool judge(){
10 int i=0;
11 for(int j=1;j<=6;j++){
12 if(a[j][1]==a[j][2]&&a[j][2]==a[j][3]&&a[j][3]==a[j][4]){
13 i++;
14 }
15 }
16 if(i==6){
17 return true;
18 }else{
19 return false;
20 }
21 }
22
23 void opera1(){//前面顺时针
24 int t1,t2;
25 t1=a[6][3];
26 t2=a[6][4];
27
28 a[6][3]=a[1][3];
29 a[6][4]=a[1][4];
30
31 a[1][3]=a[5][3];
32 a[1][4]=a[5][4];
33
34 a[5][3]=a[3][2];
35 a[5][4]=a[3][1];
36
37 a[3][1]=t2;
38 a[3][2]=t1;
39 }
40 void opera2(){//前面逆时针
41 int t1,t2;
42 t1=a[5][3];
43 t2=a[5][4];
44
45 a[5][3]=a[1][3];
46 a[5][4]=a[1][4];
47
48 a[1][3]=a[6][3];
49 a[1][4]=a[6][4];
50
51 a[6][3]=a[3][2];
52 a[6][4]=a[3][1];
53
54 a[3][1]=t2;
55 a[3][2]=t1;
56 }
57 void opera3(){//上面顺时针
58 int t1,t2;
59 t1=a[2][1];
60 t2=a[2][2];
61
62 a[2][1]=a[6][3];
63 a[2][2]=a[6][1];
64
65 a[6][3]=a[4][4];
66 a[6][1]=a[4][3];
67
68 a[4][4]=a[5][2];
69 a[4][3]=a[5][4];
70
71 a[5][2]=t1;
72 a[5][4]=t2;
73 }
74 void opera4(){//上面逆时针
75 int t1,t2;
76 t1=a[2][1];
77 t2=a[2][2];
78
79 a[2][1]=a[5][2];
80 a[2][2]=a[5][4];
81
82 a[5][2]=a[4][4];
83 a[5][4]=a[4][3];
84
85 a[4][4]=a[6][3];
86 a[4][3]=a[6][1];
87
88 a[6][1]=t2;
89 a[6][3]=t1;
90 }
91 void opera5(){//左边顺指针
92 int t1,t2;
93 t1=a[1][1];
94 t2=a[1][3];
95
96 a[1][1]=a[4][1];
97 a[1][3]=a[4][3];
98
99 a[4][1]=a[3][1];
100 a[4][3]=a[3][3];
101
102 a[3][1]=a[2][1];
103 a[3][3]=a[2][3];
104
105 a[2][1]=t1;
106 a[2][3]=t2;
107 }
108 void opera6(){//左边逆时针
109 int t1,t2;
110 t1=a[1][1];
111 t2=a[1][3];
112
113 a[1][1]=a[2][1];
114 a[1][3]=a[2][3];
115
116 a[2][1]=a[3][1];
117 a[2][3]=a[3][3];
118
119 a[3][1]=a[4][1];
120 a[3][3]=a[4][3];
121
122 a[4][1]=t1;
123 a[4][3]=t2;
124 }
125 int main()
126 {
127 int n;
128 scanf("%d",&n);
129 while(n--){
130 for(int i=1;i<=6;i++){
131 for(int j=1;j<=4;j++){
132 scanf("%d",&a[i][j]);
133 b[i][j]=a[i][j];
134 }
135 }
136 if(judge()){
137 printf("YES\n");
138 continue;
139 }
140 opera1();
141 if(judge()){
142 printf("YES\n");
143 continue;
144 }
145 for(int i=1;i<=6;i++){
146 for(int j=1;j<=4;j++){
147 a[i][j]=b[i][j];
148 }
149 }
150 opera2();
151 if(judge()){
152 printf("YES\n");
153 continue;
154 }
155 for(int i=1;i<=6;i++){
156 for(int j=1;j<=4;j++){
157 a[i][j]=b[i][j];
158 }
159 }
160 opera3();
161 if(judge()){
162 printf("YES\n");
163 continue;
164 }
165 for(int i=1;i<=6;i++){
166 for(int j=1;j<=4;j++){
167 a[i][j]=b[i][j];
168 }
169 }
170 opera4();
171 if(judge()){
172 printf("YES\n");
173 continue;
174 }
175 for(int i=1;i<=6;i++){
176 for(int j=1;j<=4;j++){
177 a[i][j]=b[i][j];
178 }
179 }
180 opera5();
181 if(judge()){
182 printf("YES\n");
183 continue;
184 }
185 for(int i=1;i<=6;i++){
186 for(int j=1;j<=4;j++){
187 a[i][j]=b[i][j];
188 }
189 }
190 opera6();
191 if(judge()){
192 printf("YES\n");
193 continue;
194 }
195 for(int i=1;i<=6;i++){
196 for(int j=1;j<=4;j++){
197 a[i][j]=b[i][j];
198 }
199 }
200 printf("NO\n");
201 }
202 return 0;
203 }
C、Pocky_hdu5984
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 54 Accepted Submission(s): 20
Problem Description
Let’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful decorative stripes in the coating, of length L.
While the length of remaining pocky is longer than d, we perform the following procedure. We break the pocky at any point on it in an equal possibility and this will divide the remaining pocky into two parts. Take the left part and eat it. When it is not longer than d, we do not repeat this procedure.
Now we want to know the expected number of times we should repeat the procedure above. Round it to 6 decimal places behind the decimal point.
Input
The first line of input contains an integer N which is the number of test cases. Each of the N lines contains two float-numbers L and d respectively with at most 5 decimal places behind the decimal point where 1 ≤ d, L ≤ 150.
Output
For each test case, output the expected number of times rounded to 6 decimal places behind the decimal point in a line.
Sample Input
6
1.0 1.0
2.0 1.0
4.0 1.0
8.0 1.0
16.0 1.0
7.00 3.00
Sample Output
0.000000
1.693147
2.386294
3.079442
3.772589
1.847298
Source
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jiangzijing2015 | We have carefully selected several similar problems for you:
5994 5993 5992 5991 5990 思路：说实话比赛的时候根本不知道怎么做出来的。
好久之后才队友想起来，输出了一下log2，突然发现了规律。
赛后看别的队，一看0.693147就知道log2了，这就是做题多了有经验了啊。
但当时说的题解好像是求微积分，还是求导呢，把公式推导出来的。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4
5 using namespace std;
6
7 int main()
8 {
9 int n;
10 double a,b;
11 scanf("%d",&n);
12 while(n--){
13 scanf("%lf%lf",&a,&b);
14 if(a<=b){
15 printf("0.000000\n");
16 continue;
17 }
18 printf("%.6lf\n",log(a)-log(b)+1);
19 }
20 return 0;
21 }
