192 条题解
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4
WiFi LV 8 @ 2017-09-02 19:24:33
stl大法好
应该很容易看懂吧
直接上代码
#include<iostream> #include<cstdio> #include<map> #include<algorithm> #include<cstring> using namespace std; int n; struct node{ string s; int x,y; }a[12]; map <string,int> q; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ cin>>a[i].s; q[a[i].s]=i; } for(int i=1;i<=n;i++){ string s1,s2; int m,k; cin>>s1; scanf("%d%d",&m,&k); if(k==0) continue; a[q[s1]].x+=m/k*k; m/=k; for(int j=1;j<=k;j++){ cin>>s2; a[q[s2]].y+=m; } } for(int i=1;i<=n;i++){ cout<<a[i].s<<" "; printf("%d\n",a[i].y-a[i].x); } return 0; } -
1
/* 此题简单模拟，但也有细节要注意： 1.人数有可能为0，在处理剩余钱时要判断人数是否为0， 不然会出现除以0的情况 2.老问题了，可能是做题少，双重循环的自加条件， 总是容易将内循环自加写成外循环自加 3.注意num数组的找分钱人时的方法 总而言之，虽然这类题目简单，但竟然还做了很久， 应该多做点题锻炼一下基础能力了，不能贪图高级算法了 */ #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> using namespace std; struct node { string name; int out,friends,in; int num[11]; }a[12]; int n; int find(string b) { for(int i=1;i<=n;i++) if(a[i].name==b) return i; return 0; } int main() { cin>>n; for(int i=1;i<=n;i++) cin>>a[i].name; string b; for(int i=1;i<=n;i++) { cin>>b; int x=find(b); cin>>a[x].out>>a[x].friends; for(int j=1;j<=a[x].friends;j++) { cin>>b; a[x].num[j]=find(b); } } for(int i=1;i<=n;i++) { if(a[i].friends==0) continue; int k=a[i].out/a[i].friends; a[i].out=k*a[i].friends; for(int j=1;j<=a[i].friends;j++) a[a[i].num[j]].in+=k; } for(int i=1;i<=n;i++) cout<<a[i].name<<" "<<a[i].in-a[i].out<<endl; return 0; } -
0
#include<iostream> #include<map> #define FOR(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int N=20; map<string,int>a; map<string,int>::iterator it; int n,m,p; string s,name[12]; int main() { cin>>n;FOR(i,1,n)cin>>name[i]; FOR(i,1,n) { cin>>s>>m>>p; int v; if(p==0)continue; v=m/p; a[s]-=v*p; FOR(j,1,p) { cin>>s; a[s]+=v; } } FOR(i,1,n) cout<<name[i]<<" "<<a[name[i]]<<endl; return 0; } -
0
#include <iostream> //[USACO]贪婪的送礼者 #include <algorithm> #include <string> #include <unordered_map> using namespace std; int main() { unordered_map<string, int> m; string str; int NP, tatol, n; cin >> NP; string N[10]; for (int i = 0; i < NP; i++) { cin >> str; N[i] = str; m[str] = 0; } for (int i = 0; i < NP; i++) { cin >> str >> tatol >> n; int k; if(n) k = tatol / n; m[str] -= k * n; for (int i = 0; i < n; i++) { cin >> str; m[str] += k; } } for (int i = 0; i < NP; i++) cout << N[i] << " " << m[N[i]] << endl; return 0; } -
0
之前用c语言写不知道为什么一直不行，受到大佬启发，用map试了一下。
可以说是本弱智儿童做出第一道题。
一下是我的代码。
#include<iostream>
#include<map>
#include<cstring>
using namespace std;
int main()
{
int money, rest, k, num;
char name[10][20];
map<string, int> p;
cin >> num;
for (int i = 0; i < num; i++) {
cin >> name[i];
p.insert(pair<string,int>(name[i],0));
}
for (int i = 0; i < num; i++) {
char receive_name[20];
char temp_name[20];
cin >> temp_name;
cin >> money >> k;
if (k == 0) continue;
else {
rest = money - (money / k)*k;
}
for (int j = 0; j < k; j++) {
cin >> receive_name;
p[receive_name] += (money / k);
}
p[temp_name] += rest;
p[temp_name] -= money;
}
for (int i = 0; i < num; i++)
cout << name[i] << " " << p[name[i]]<<endl;
return 0;
} -
0
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; int n; struct peo { char name[30]; int money; int income; }; void solve() { peo list[50]; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%s",list[i].name); list[i].money=0; list[i].income=0; } for(int i=0;i<n;i++) { int num,sig,pos; char tmpname[30]; scanf("%s",tmpname); for(pos=0;pos<n;pos++) { if(strcmp(tmpname,list[pos].name)==0) { break; } } scanf("%d %d",&list[pos].money,&num); if(num==0) { continue; } else { sig=list[pos].money/num; } list[pos].income+=list[pos].money-sig*num; for(int j=0;j<num;j++) { char str[30]; int lp; scanf("%s",str); for(lp=0;lp<n;lp++) { if(strcmp(str,list[lp].name)==0) { break; } } list[lp].income+=sig; } } for(int i=0;i<n;i++) { printf("%s %d\n",list[i].name,list[i].income-list[i].money); } } int main() { solve(); return 0; } -
0
wangqihang LV 8 @ 2015-02-28 10:17:22
太简单了，秒杀这题。。。。。。 哈
program e1035;
var n,i,j,x,y,t,k,b:longint;
d,e,f:array[1..100000]of longint;
a:array[1..10000]of string; s,c:string;
begin
// assign(input,'1.txt');
// reset(input);
readln(n);
for i:=1 to n do
readln(a[i]);
for i:=1 to n do
begin
readln(s);
for j:=1 to n do
if s=a[j] then
begin
t:=j;
break;
end;
readln(b,k);
if k<>0 then
begin
for j:=1 to n do
if s=a[j] then
begin
f[t]:=f[t]-b+b mod k;
break;
end;
for j:=1 to k do
begin
readln(s);
for x:=1 to n do
if s=a[x] then
begin
f[x]:=f[x]+b div k;
break;
end;
end;
end;
end;
for i:=1 to n do
writeln(a[i],' ',f[i]);
end. -
0
program e1035;
var n,i,j,x,y,t,k,b:longint;
d,e,f:array[1..100000]of longint;
a:array[1..10000]of string; s,c:string;
begin
// assign(input,'1.txt');
// reset(input);
readln(n);
for i:=1 to n do
readln(a[i]);
for i:=1 to n do
begin
readln(s);
for j:=1 to n do
if s=a[j] then
begin
t:=j;
break;
end;
readln(b,k);
if k<>0 then
begin
for j:=1 to n do
if s=a[j] then
begin
f[t]:=f[t]-b+b mod k;
break;
end;
for j:=1 to k do
begin
readln(s);
for x:=1 to n do
if s=a[x] then
begin
f[x]:=f[x]+b div k;
break;
end;
end;
end;
end;
for i:=1 to n do
writeln(a[i],' ',f[i]);
end. -
0
我擦，又是这样，不好好看题目，没看到还有保留钱，没一次AC，丢脸了。这都多少次了，诶，下次一定要耐心看题目！！！！！！！
总之不废话，简单的模拟即可，自己别看晕了就好，字符串程序的读入有点恶心。
###Bolck code
program P1035;
var a,np,i,num,j,np2:longint; sentter,getter:string;
money:array[1..10] of longint;
names:array[1..10] of string;
function find(x:string):longint;
var i:longint;
begin
for i:=1 to np do
if x=names[i] then
exit(i);
end;begin //main
//assign(input,'shuju.in'); reset(input);readln(np); fillchar(money,sizeof(money),0);
for i:=1 to np do readln(names[i]);for i:=1 to np do
begin
readln(sentter); read(num); readln(np2);
a:=find(sentter);money[a]:=money[a]-num;
if np2<>0 then
money[a]:=money[a]+(num mod np2);for j:=1 to np2 do
begin
readln(getter); a:=find(getter); money[a]:=money[a]+(num div np2);
end;
end;for i:=1 to np do
begin
write(names[i],' '); writeln(money[i]);
end;
//close(input);
end.-
冲啊小笼包 @ 2015-01-09 18:58:19
注解：find函数用来搜索这个人是第几个人在数据，因为发礼物的时候顺序不一样- -当然你也可以离线处理，随便啦反正就10个人
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-
0
#include<stdio.h>
#include <string.h>struct student
{char name[14];
int h_money;
int s_money;
int g_money;
}a[10];int main()
{
int n,i,j,m,k,r;
int result;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",a[i].name);
}
for(i=0;i<n;i++)
{
char temp_name[15]={0};
scanf("%s",temp_name);
for(j=0;j<n;j++)
{
if (strcmp(a[j].name,temp_name)==0)
{
scanf("%d%d",&a[j].h_money,&m);
for(k=0;k<m;k++)
{
char g_money_name[15]={0};
scanf("%s",g_money_name);
for(r=0;r<n;r++)
{
if(strcmp(a[r].name,g_money_name)==0)
{
a[r].g_money=a[r].g_money+a[j].h_money/m;
a[j].s_money=a[j].s_money+a[j].h_money/m;
}
}
}
}
}
}for(i=0;i<n;i++)
{
result=a[i].g_money-a[i].s_money;
printf("%s %d\n",a[i].name,result);
}return 0;
} -
0
NOIP2014赛前AC留念
var np,i,j,z,tip:longint;
t:string;
s,give:array[1..10] of string;
peo,mon,num:array[1..50] of longint;begin
//assign(input,'gift1.in');
//assign(output,'gift1.out');
//reset(input);
//rewrite(output);
readln(np);
for i:=1 to np do
readln(s[i]);
for i:=1 to np do
begin
readln(t);
for j:=1 to np do
if t=s[j] then
begin
tip:=j;
break;
end;readln(mon[tip],peo[tip]);
if peo[tip]<>0 then
begin
num[tip]:=num[tip]-mon[tip]+(mon[tip] mod peo[tip]);
for j:=1 to peo[tip] do
begin
readln(give[j]);
for z:=1 to np do
if s[z]=give[j] then
begin
num[z]:=num[z]+(mon[tip] div peo[tip]);
break;
end;
end;
end;
end;
for i:=1 to np do
writeln(s[i],' ',num[i]);
//close(input);
//close(output);
end. -
0
#include<cstdio>
#include<cstring>using namespace std;
struct Person_node
{
char name[15];
int Send_money,Receive_money,Receive_num;
}Person[11];int N;
int search(char *name)
{
for(int i=1;i<=N;i++)
if (strcmp(name,Person[i].name)==0)
return i;
}int main()
{
memset(Person,0,sizeof(Person));
scanf("%d\n",&N);
for(int i=1;i<=N;i++)
scanf("%s\n",Person[i].name);
char name[15];
for(int i=1;i<=N;i++)
{
scanf("%s\n",name);
int id=search(name);
scanf("%d\n%d\n",&Person[id].Send_money,&Person[id].Receive_num);
if (Person[id].Send_money!=0)
Person[id].Send_money-=Person[id].Send_money%Person[id].Receive_num;
for(int j=1;j<=Person[id].Receive_num;j++)
{
scanf("%s\n",name);
Person[search(name)].Receive_money+=Person[id].Send_money/Person[id].Receive_num;
}
}
for(int i=1;i<=N;i++)
printf("%s %d\n",Person[i].name,Person[i].Receive_money-Person[i].Send_money);
return 0;
} -
0
2691453579 LV 4 @ 2014-09-07 14:23:58
type rec=record
a:string;
b:longint;
c:longint;
end;
var n,i,j,num,l,give:longint;
s1,s2:string;
s:array[1..10] of rec;
begin
readln(n);
for i:=1 to n do readln(s[i].a);
for l:=1 to n do begin
readln(s1);
give:=0;
for j:=1 to n do begin
if s[j].a=s1 then begin
readln(s[j].b,num);
if num=0 then break;
give:=s[j].b div num;
s[j].b:=give*num;
end;
end;
for i:=1 to num do begin
readln(s2);
for j:=1 to n do begin
if s[j].a=s2 then inc(s[j].c,give);
end;
end;
end;
for i:=1 to n do begin
writeln(s[i].a,' ',s[i].c-s[i].b);
end;
end. -
0
type
node=record
name:string;
min,mout:integer;
end;
var
p:array[1..10]of node;
n,m,x,i,j,k,t:integer;
name,namem:string;
begin
readln(n);
for i:=1to n do
readln(p[i].name);
for i:=1to n do
begin
readln(namem);
for j:=1to n do if p[j].name=namem then begin
t:=j;
break;
end;
readln(m,x);
for j:=1to x do
begin
readln(name);
for k:=1to n do
if p[k].name=name then begin
inc(p[k].min,m div x);
break;
end;
end;
if x<>0then inc(p[t].mout,m-m mod x);
end;
for i:=1to n do
writeln(p[i].name,' ',p[i].min-p[i].mout);
end.
发代码老是格式错误挤成一堆 所以我只好用打一行空一行这种恶心的办法==题目不难纯模拟但是要细心点……坑了我半天
-
0
20题！！！
记录信息
评测状态 Accepted
题目 P1035 贪婪的送礼者
递交时间 2014-08-02 16:23:53
代码语言 C++
评测机 VijosEx
消耗时间 15 ms
消耗内存 272 KiB
评测时间 2014-08-02 16:23:58
评测结果
编译成功测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #2: Accepted, time = 15 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 268 KiB, score = 10
Accepted, time = 15 ms, mem = 272 KiB, score = 100 -
0
Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步 -
0
Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步 -
0
var
nam:array[0..11] of string;
res:array[0..11] of longint;
n,ger,rer,i,j,m,pay:longint;Function num:longint;
begin num:=1; while (nam[num]<>nam[0]) do inc(num);
end;begin
readln(n);
for i:=1 to n do readln(nam[i]);
for i:=1 to n do begin
readln(nam[0]); read(pay); readln(m);
ger:=num; if m=0 then continue;
res[ger]:=res[ger]-pay+pay mod m;
pay:=pay div m;
for j:=1 to m do begin
readln(nam[0]); rer:=num;
res[rer]:=res[rer]+pay;
end;
end;
for i:=1 to n do writeln(nam[i],' ',res[i]);
end.
WiFi
LV 8
@ 2017-09-02 19:24:33
