279 条题解
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0
yangjz1125 LV 7 @ 2018-08-26 13:51:01
nowm = 0 savem = 0 for i in range(1,13): tmp = int(input()) nowm += 300 if tmp > nowm: print("-{}".format(i)) exit() else: nowm -= tmp savem += (nowm // 100) nowm %= 100 nowm += savem * 120 print(nowm)python 就是简洁
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0
周煊博 (bill2007) LV 8 @ 2018-08-18 18:01:25
p党标程
var
i,s,x,bank:longint;
begin
for i:=1 to 12 do
begin
s:=s+300;
readln(x);
if s<x then
begin
writeln(-i);
halt;
end;
s:=s-x;
bank:=bank+s div 100;
s:=s mod 100;
end;
writeln(bank*120+s);
end. -
0
绝对的一A
#include<iostream> using namespace std; int main() { int y=0,ans=0,s=0,n,c=0; bool a=1; for(int i=1;i<=12;i++) { cin>>n; y=s+300; if(a&&y<n) { ans=i-i-i; a=0; } else { c+=(y-n)/100*100; s=(y-n)%100; y-=(y-n)/100*100; y-=n; } } if(a) cout<<c/100*120+s; else cout<<ans; return 0; } -
0
#include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> using namespace std; int main() { int i=0,x,y=0,a; float t=0; while(i<12) { ++i; cin>>x; if(300+y-x<0) { cout<<"-"<<i; return 0; } a=(300+y-x)/100; t=t+1.2*100*a; y=300-100*a+y-x; } cout<<t+y; return 0; } -
0
【分析】
此题考查的是模拟:
1）框架
int main()
{
int 定义：结余，预算，存款；
for (从1月~12月)
{
输入预算；
得出本月结余并存款；
当结余出现负数时：
}
输出总的攒钱数；
return 0;
}
2)运行过程当中（样例），可能会输入一半就输出-7了，很正常。若想全部输出，可以利用文件的输入与输出，这里介绍一下文件的处理方法：
[注意：双斜线中的内容]
#include <cstdio> //文件所用到的头文件cstdio
using namespace std;
int main()
{
freopen("a+b=c.in","r",stdin); //将输入文件开
freopen("a+b=c.out","w",stdout);//将输出文件开
int a,b,c;
scanf("%d%d",&a,&b);
c=a+b; //操作
printf("%d\n",c);
fclose(stdin);fclose(stdout); //将输入、输出文件关
return 0; //结束程序
}
【标程】
#include <iostream>
#define MONTH 300
using namespace std;
int main()
{
int change=0,cost,bank=0;//结余，预算，存款
for (int i=1;i<=12;i++)//从1月~12月
{
cin>>cost;//输入
change+=MONTH-cost;//得出本月结余
if (change>=100)//如果本月结余可以存款
{
bank+=change/100*100;//存款累计
change%=100;//求出存款后的结余
}
if(change<0)//当结余出现负数时
{
cout<<"-"<<i<<endl;//输出
return 0;
}
}
cout<<bank*(1+20*1.0/100)+change<<endl;//输出月末钱
return 0;
} -
0
等下。时光请等一下 (天下第一招3) LV 7 @ 2018-05-10 17:06:04
#include<bits/stdc++.h> using namespace std; int main(){ int arr1[13];//数组越界问题 int sum=0,a=0,c=0; for(int i=1;i<=12;i++){ cin>>arr1[i]; } for(int i=1;i<=12;i++){ if(i==1){ a=300-arr1[i];//剩余的钱 if(a<0){ cout<<"-1"<<endl; break; }else{ sum+=(a/100)*100;//存的钱 c=a-((a/100)*100);//除去存的钱之外，还剩下多少钱? } } } for(int i=2;i<=12;i++){ a=300-arr1[i]+c; if(a<0){ cout<<"-"<<i<<endl; break; }else{ sum+=(a/100)*100; c=a-((a/100)*100); } if(i==12){ cout<<sum*1.2+c<<endl; } } return 0; }这个题不是很难理解，但是一直想不通不能AC，
困扰我一两天，突然突发奇想，是不是数组的空间不够啊，
改了下，果然成功了，心态崩了 -
0
beyond2004 LV 5 @ 2018-04-29 15:49:49
简单C++程序
#include<iostream>
using namespace std;
int main(){
int c,s=0,h=0;
for(int i=1;i<=12;i++)
{
cin>>c;
s-=c-300;
if(s<0)
{
cout<<"-"<<i;
return 0;
}
else h+=s/100,s%=100;
}
cout<<(120*h+s);
return 0;
}-
nbnb
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0
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int a[13],s,ans=0,k=0,j=0;
float x;
for (int i=1;i<=12;i++)
{
scanf("%d",&a[i]);
s=300+ans-a[i];
if (s<0)
{
printf("%d",-i);
return 0;
}
j=s;
s/=100;
k+=s*100;
j-=s*100;
ans=j;
}
x=k*1.2;
ans+=x;
printf("%d",ans);
return 0;
} -
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feidaoluoye LV 9 @ 2017-11-21 16:36:07
#include <iostream> #include<cstdlib> #include<cstdio> #include<map> #include<vector> #include<cstring> #include<algorithm> #define mod 7654321 #define FOR(i,x,y) for(i=x;i<=y;++i) #define maxa 1000+100 using namespace std; int a[maxa]; int main() { int i; FOR(i,1,12) cin>>a[i]; int t,m,s = 0; t =0; FOR(i,1,12) { s+=300; if(s<a[i]) { cout<<"-"<<i<<endl; return 0; } else { s-=a[i]; t+=s/100; s = s%100; } } cout<<t*120+s<<endl; } -
0
#include<iostream>
int main()
{
using namespace std;
int sum=0,x=0,y,a[13];
for (int i=1;i<=12;i++) cin>>a[i];
for (int i=1;i<=12;i++)
{
sum+=300;
if (sum<a[i])
{
cout<<-i;return 0;
}
while (sum-a[i]>=100)
{
sum-=100;
y+=100;
}
sum-=a[i];
}
cout<<sum+y*1.2<<endl;
return 0;
}
c++新人，，，刚开始用pascal提交，好无语 -
0
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#define pi 3.14159265358979
#define eps 1e-10
#define TN 50002
using namespace std;
int a[101];
int ans,tt,l;
int main()
{
for(int i=1;i<=12;i++) scanf("%d",&a[i]);
for(int i=1;i<=12;i++)
{
ans+=300;
if(ans<a[i]) {printf("-%d",i);return 0;}//if(ans-a[i]>=100)
{l=(ans-a[i])/100;ans=(ans-a[i])%100;tt+=l*100;}
}
printf("%d",120*tt/100+ans);
return 0;
} -
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shi.shu.huan LV 8 @ 2017-08-23 19:01:14
#include<cmath>
#include<iostream>
using namespace std;
int cant=0;
double mid;
double ans;
int ys[12],nd;
int own,mom;
int main()
{
for(int i=1;i<=12;i++)
cin>>ys[i-1];
for(int i=1;i<=12;i++)
{
own=own-ys[i-1];
if(own<0)
{
nd=-own;
mid=(double)nd/100;
if(floor(mid)==mid)
nd=(floor(mid))*100;
else
nd=(floor(mid)+1)*100;
if(nd>300)
{
cant=1;
ans=-i;
break;
}}
own=own+nd;
mom=mom-nd+300;
nd=0;
}
if(cant==0)
ans=(double)own+(mom*1.2);
cout<<ans;
return 0;
} -
0
真的很简单……
include<iostream>
#include<cstdio>
using namespace std;
int main()
{
long long a[13],m=1,n,sum=0,x,f;
double v=0;
for(int i=1;i<=12;++i)
{
cin>>a[i];
sum=sum+300-a[i];
if(sum<0)
{
m=0;
f=i;
break;
}
else
{
n=sum/100;
v=v+n*100;
sum=sum%100;
}
}
if(m==1)
{
v=v*1.2+sum;
printf("%.0f",v);
}
else
{
cout<<"-"<<f;
}
} -
0
#include <iostream>
using namespace std;
int main(){
int money=0;
int count=0;
int j;
for( j=1;j<=12;j++){
int cost;
cin>>cost;
money+=300;
if(money<cost){
cout<<-j<<endl;
break;
}
else {
int n=(money-cost)/100;
count+=n;
money=money-cost-n*100;
}
}
if(j==13)cout<<(money+1.2*count*100)<<endl;
return 0;
} -
0
WHAT A PITY!
#include<stdio.h>
int main()
{
int month[13]={0};
int j=0,mom=0,i;
float all;
for(i=1;i<13;i++)
scanf("%d",&month[i]);
for(i=1;i<13;i++)
{
j=300-month[i];
if(j<0)
{
printf("-%d",i);
return 0;
}
mom=j/100*100;
j=j%100;
}
all=mom*1.2+j;
printf("%g",all);
}
yangjz1125
LV 7
@ 2018-08-26 13:51:01
