string和数值之间的转换（均为C++11）
to_string(val) //可将任何算术类型转为数值val的string表示
stoi(s,p,b) /*返回s的起始子串（表示整数内容）的数值，b表示转换所用的 基数，默认值为10；p是size_t指针，用来保存s中第一个非数值 字符的下标，默认为0，即不保存下标 每个类型都有相应的 to_string版本*/
//返回int
stol(s,p,b) //返回long
stoul(s,p,b) //返回unsigned long
stoll(s,p,b) //返回long long
stoull(s,p,b) //返回unsigned long long
stof(s,p) /*返回s的起始子串（表示浮点数内容）的数值 //返回float
stod(s,p) //返回double
stold(s,p) //返回long double 下面为代码
1001 A+B Format
#include<iostream>
#include<string>
#include<stdio.h>
using namespace std;
int main(){
int a,b,sum;
char p[10]={0};
string s;
cin>>a>>b;
sum=a+b;
// s=to_string(static_cast<long long>(sum)); 使用to_string (若在支持C++11特性的情况下可直接用)
sprintf(p,"%d",sum);
s=p;
int temp=0; //temp用来判断为负数还是正数
if(*s.begin()=='-'){
temp=1;
}
for(int i=s.size()-3;i>temp;i=i-3){
s.insert(i,1,',');
}
cout<<s<<endl;
return 0;
}
1002 A+B for Polynomials（多项式）
题目大意为 输入两行多项式，相加
Input
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents（指数） and coefficients（系数）, respectively. It is given that 1 <= K <= 10，0 <= NK < … < N2 < N1 <=1000.
Output
For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
Sample Input
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output
3 2 1.5 1 2.9 0 3.2
以下为我的代码
最后一个测试点有一个段错误。。很尴尬，，应该是vector的insert的问题，但是由于我接触vector不太久，所以没找到。。。以后找到了更新
#include<iostream>
#include<vector>
using namespace std;
class Polynomial{
private :
int exponent;
float coefficient;
public:
void setexponent(int exponent){
this->exponent=exponent;
}
void setcoeffient(float coefficient){
this->coefficient=coefficient;
}
int getexponent(){
return this->exponent;
}
float getcoeffient(){
return this->coefficient;
}
};
int main(){
vector<Polynomial> po1;
vector<Polynomial> po2;
vector<Polynomial> sum;
int K,exponent;
float coefficient;
for(int i=0;i<=1;++i){
cin>>K;
for(int j=0;j<K;++j){
cin>>exponent>>coefficient;
Polynomial s;
s.setexponent(exponent);
s.setcoeffient(coefficient);
if(i==0){
po1.push_back(s);
}else{
po2.push_back(s);
}
}
}
auto begin1=po1.begin();
auto begin2=po2.begin();
for(;begin1!=po1.end()&&begin2!=po2.end();){
if((*begin1).getexponent()>(*begin2).getexponent()){
sum.push_back(*begin1);
++begin1;
}else if((*begin1).getexponent()<(*begin2).getexponent()){
sum.push_back(*begin2);
++begin2;
}else{
float x=(*begin1).getcoeffient()+(*begin2).getcoeffient();
if(x){
Polynomial s;
s.setexponent((*begin1).getexponent());
s.setcoeffient(x);
sum.push_back(s);
}
++begin1;
++begin2;
}
}
if(begin1!=po1.end()){
sum.insert(sum.end(),begin1,po1.end());
}else if(begin2!=po2.end()){
sum.insert(sum.end(),begin2,po2.end());
}
cout<<sum.size()<<" ";
auto begin=sum.begin();
for(;begin!=sum.end()-1;++begin){
printf("%d %0.1f ",(*begin).getexponent(),(*begin).getcoeffient());
}
printf("%d %0.1f",(*begin).getexponent(),(*begin).getcoeffient());
return 0;
}
