A模拟
只需要对项目a和项目b进行选择
假设此时为t秒，选择项目a和b
如果先选项目a
那么时间为
\(a.a+a.b*t+(a.a+a.b*t+t)*b.b + b.a\)
如果先选项目b
那么时间为
\(b.a+b.b*t+(b.a+b.b*t+t)*a.b+a.a\)
相约之后为
\(a.a*b.b\)和\(b,a*a.b\)
那么按照这个排序即可
注意开long long即可
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#define ll long long
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;
const ll mod = 365 * 24 * 60 * 60;
using namespace std;
struct node{
ll a,b;
}e[maxn];
bool cmp(node a,node b){
return a.a * b.b < b.a * a.b;
}
int main(){
int n;
while(~scanf("%d",&n)){
if(n==0)break;
for(int i=0;i<n;i++)scanf("%lld%lld",&e[i].a,&e[i].b);
sort(e,e+n,cmp);
ll ans = 0;
for(int i=0;i<n;i++){
ans = (ans * e[i].b + e[i].a + ans) % mod;
}
printf("%lld\n",ans);
}
return 0;
}I水题
求出所有数字的平方和即可
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;
using namespace std;
int main(){
int n;
while(cin>>n){
if(n==0)break;
int sum = 0;
int x;
for(int i=0;i<n;i++){
cin>>x;
sum += pow(x,2);
}
cout<<sum<<endl;
}
return 0;
}J组合问题
n件衣服，m条裤子，k双鞋子
给出p个要求，则只能进行这样搭配
题意可知裤子是中间量，只需要知道对于每一条裤子，有多少种衣服和鞋子可以搭配即可
对于给出衣服和裤子
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#define ll long long
using namespace std;
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
int n,m,k;
int getans(int p){
char s1[10],s2[10];
int a,b;
int clothes[1005],shoes[1005];
memset(clothes,0,sizeof(clothes));
memset(shoes,0,sizeof(shoes));
while(p--){
cin>>s1>>a>>s2>>b;
if(!strcmp(s1,"clothes")){
clothes[b]++;
}else{
shoes[a]++;
}
}
int ans = 0;
for(int i=1;i<=m;i++){
ans += (n - clothes[i]) * (k - shoes[i]);
}
return ans;
}
int main(){
while(~scanf("%d%d%d",&n,&m,&k)){
if(n==0&&m==0&&k==0)break;
int p;
cin>>p;
int ans = 0;
if(p==0)ans = n*m*k;
else ans = getans(p);
cout<<ans<<endl;
}
return 0;
}K超级大模拟
耐心算出规律即可
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;
using namespace std;
struct node{
int s;//速度
int t;//转弯的时间
char c;//方向
int x,y;//坐标
}tom,jerry;
int k;
int n;
void print(){
printf("%d %d\n",tom.x,tom.y);
printf("%d %d\n",jerry.x,jerry.y);
}
void change(node &a,int i){//检测是否左转
if(i % a.t == 0){
switch(a.c){
case 'W':a.c = 'S';break;
case 'E':a.c = 'N';break;
case 'S':a.c = 'E';break;
case 'N':a.c = 'W';break;
}
}
}
bool check(){//检测是否需位于同一个点且交换方向
if(tom.x == jerry.x && tom.y == jerry.y){
swap(tom.c,jerry.c);
return true;
}
return false;
}
void move(node &a){
if(a.c == 'W'){//左前进
if(a.y > a.s){
a.y -= a.s;
}else if(a.y <= a.s && a.s <= n-1+a.y){
a.y = a.s - (a.y - 2);
a.c = 'E';
}else{
a.y = a.s - a.y;
}
}else if(a.c == 'E'){//右前进
if(a.y + a.s <= n){
a.y += a.s;
}else if(a.s >= n-a.y && a.s <= n-a.y+n-1){
a.y = 2 * n - a.y - a.s;
a.c = 'W';
}else{
a.y = a.s - (2*(n-1)-a.y);
}
}else if(a.c == 'N'){//上前进
if(a.x > a.s){//不换方向
a.x -= a.s;
}else if(a.x <= a.s && a.s <= n-1+a.x){//只换一次方向
a.x = a.s - (a.x - 2);
a.c = 'S';
}else{//换两次方向
a.x = a.s - a.x;
}
}else if(a.c == 'S'){//下前进
if(a.x + a.s <= n){
a.x += a.s;
}else if(a.s >= n-a.x && a.s <= n-a.x+n-1){
a.x = 2 * n - a.x - a.s;
a.c = 'N';
}else{
a.x = a.s - (2*(n-1)-a.x);
}
}
}
void getans(){//先运到，判断是否在同一点然后交换方向，再判断转弯
for(int i=1;i<=k;i++){
move(tom);
move(jerry);
bool xx = check();
if(xx==true)continue;
change(tom,i);
change(jerry,i);
}
print();
}
void init(){//初始位置和削弱速度
tom.x = tom.y = 1;
jerry.x = jerry.y = n;
tom.t = tom.t % (2 * (n - 1));
jerry.t = jerry.t % (2 * (n - 1));
}
int main(){
while(cin>>n){
if(n==0)break;
cin>>tom.c>>tom.s>>tom.t;
cin>>jerry.c>>jerry.s>>jerry.t;
init();
cin>>k;
getans();
}
return 0;
}