1104. Sum of Number Segments (20)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CAO, Peng
Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence {0.1, 0.2, 0.3, 0.4}, we have 10 segments: (0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) (0.4).
Now given a sequence, you are supposed to find the sum of all the numbers in all the segments. For the previous example, the sum of all the 10 segments is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N, the size of the sequence which is no more than 105. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.
Output Specification:
For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.
Sample Input:
4
0.1 0.2 0.3 0.4
Sample Output:
5.00
代码如下：
注意点：用double不要用float
#include<cstdio>
#include<iostream>
using namespace std;
int main(){
int n;
scanf("%d",&n);
double value;
double sum=0.00;
for(int i=0;i<n;++i){
cin>>value;
if(i==0||i==n-1){
sum+=value*n;
}else{
sum+=value*(n-i)*(i+1);
}
}
printf("%0.2f",sum);
return 0;
}
1105. Spiral Matrix (25)
时间限制
150 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 104. The numbers in a line are separated by spaces.
Output Specification:
For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.
Sample Input:
12
37 76 20 98 76 42 53 95 60 81 58 93
Sample Output:
98 95 93
42 37 81
53 20 76
58 60 76
此题题意为螺旋复制
有四步，先横再竖再横再竖，一直循环，知道所有空位被赋值
代码如下：
#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define max 10001
int matrix[max][max];
bool cmp1(int a,int b){
return a>b;
}
int main(){
int N,m,n;
vector<int> vec;
scanf("%d",&N);
int value;
for(int i=0;i<N;++i){
scanf("%d",&value);
vec.push_back(value);
}
sort(vec.begin(),vec.end(),cmp1);
n=0;
for(int i=1;i*i<=N;++i){
if(N%i==0&&i>n){
n=i;
}
}
m=N/n;
int i=0,j=0;
int n1=0;
int m1=0;
int n2=n;
int m2=m;
int index=0;
while(index<N){
for(;j<n2;++j){
matrix[i][j]=vec[index++];
if(index==N)
{
break;
}
}
if(index==N)
{
break;
}
--n2;
++i;
--j;
for(;i<m2;++i){
matrix[i][j]=vec[index++];
if(index==N)
{
break;
}
}
if(index==N)
{
break;
}
--m2;
--i;
--j;
for(;j>=n1;--j){
matrix[i][j]=vec[index++];
if(index==N)
{
break;
}
}
if(index==N)
{
break;
}
++n1;
--i;
++j;
for(;i>m1;--i){
matrix[i][j]=vec[index++];
if(index==N)
{
break;
}
}
++m1;
++i;
++j;
}
for(int a=0;a<m;++a){
for(int b=0;b<n;++b){
if(b!=n-1){
printf("%d ",matrix[a][b]);
}else{
printf("%d",matrix[a][b]);
}
}
printf("\n");
}
return 0;
}
1100. Mars Numbers (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
People on Mars count their numbers with base 13:
- Zero on Earth is called “tret” on Mars.
- The numbers 1 to 12 on Earch is called “jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec” on Mars, respectively.
- For the next higher digit, Mars people name the 12 numbers as “tam, hel, maa, huh, tou, kes, hei, elo, syy, lok, mer, jou”, respectively.
For examples, the number 29 on Earth is called “hel mar” on Mars; and “elo nov” on Mars corresponds to 115 on Earth. In order to help communication between people from these two planets, you are supposed to write a program for mutual translation between Earth and Mars number systems.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (< 100). Then N lines follow, each contains a number in [0, 169), given either in the form of an Earth number, or that of Mars.
Output Specification:
For each number, print in a line the corresponding number in the other language.
Sample Input:
4
29
5
elo nov
tam
Sample Output:
hel mar
may
115
13
代码
#include<iostream>
#include<string>
#include<vector>
#include<sstream>
using namespace std;
string low[13]={"tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
string high[13]={" ","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
void todigitlow(string s){
if(s.size()==3){
for(int i=0;i<13;++i){
if(s==low[i]){
cout<<i<<endl;
return;
}else if(s==high[i]){
cout<<i*13<<endl;
return;
}
}
}else{
string s1=s.substr(0,3);
string s2=s.substr(4,3);
int sum=0;
for(int i=0;i<13;++i){
if(s1==high[i]){
sum+=i*13;
}
if(s2==low[i]){
sum+=i;
}
}
cout<<sum<<endl;
return;
}
}
string tolowmars(int value){
switch(value){
case 0:
return "tret";
case 1:
return "jan";
case 2:
return "feb";
case 3:
return "mar";
case 4:
return "apr";
case 5:
return "may";
case 6:
return "jun";
case 7:
return "jly";
case 8:
return "aug";
case 9:
return "sep";
case 10:
return "oct";
case 11:
return "nov";
case 12:
return "dec";
}
}
string tohighmars(int value){
switch(value){
case 1:
return "tam";
case 2:
return "hel";
case 3:
return "maa";
case 4:
return "huh";
case 5:
return "tou";
case 6:
return "kes";
case 7:
return "hei";
case 8:
return "elo";
case 9:
return "syy";
case 10:
return "lok";
case 11:
return "mer";
case 12:
return "jou";
default :
return "tret";
}
}
int main(){
int n;
cin>>n;
getchar(); //避免getline
string s;
for(int i=0;i<n;++i){
getline(cin,s);
if(isdigit(s[0])){
vector<string> vec;
stringstream ss;
int value;
ss<<s;
ss>>value;
if(value%13!=0){
vec.push_back(tolowmars(value%13));
}else{
cout<<tohighmars(value/13)<<endl;
continue;
}
while(value/13){
value/=13;
// cout<<tohighmars(value%13)<<" "<<value<<endl;
vec.push_back(tohighmars(value%13));
}
for(int j=vec.size()-1;j>=0;--j){
if(j!=0){
cout<<vec[j]<<" ";
}else{
cout<<vec[j]<<endl;
}
}
}else{
todigitlow(s);
}
}
return 0;
}
