浙江中医药大学第十一届程序设计竞赛题解
官方题解:http://www.jnxxhzz.com/Article/article/9.html
2019: 特产
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 548 Solved: 154
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Description
Input
Output
输出一个整数表示dd带回来的特产重量
Sample Input
2
3 6
1 3
Sample Output
3
2
【分析】:注意是实数,不要用cin会超时。
【代码】:
View Code
Submit: 591 Solved: 141
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![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; #define ll long long #define PI 3.14159 int t; int main() { scanf("%d",&t); while(t--) { double n,m; scanf("%lf%lf",&n,&m); printf("%.0f\n",(m-n)); } return 0; }
2020: Pizza
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 591 Solved: 141
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Description
Input
Output
输出cc最少会获得的卡路里
Sample Input
1
1 2
Sample Output
2
HINT
【分析】:最少那就只吃一块pizza。
【代码】:
View Code
Submit: 330 Solved: 115
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![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; #define ll long long #define PI 3.14159 int t; int n; double k; int main() { scanf("%d",&t); while(t--) { scanf("%d%lf",&n,&k); printf("%.0f\n",1.0*k); } return 0; }
不忘初心,砥砺前行!
2024: cc的神奇背包
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 330 Solved: 115
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Description
Input
Output
Sample Input
1
4 2
1 2
2 1
3 1
2 3
Sample Output
yes
【分析】:结构体排序。
【代码】:有注释。
结构体排序
Submit: 147 Solved: 26
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![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; #define ll long long #define PI 3.14159 int t; int n,k; struct node { int x,y; }a[5000]; //int a[5000],b[5000]; int cmp(node a,node b) { return a.x<b.x; //体积小的先放 return a.y>b.y; //扩容大的先放 } int f=1; int main() { scanf("%d",&t); while(t--) { f=1; scanf("%d%d",&n,&k); for(int i=0;i<n;i++) { scanf("%d%d",&a[i].x,&a[i].y); } sort(a,a+n,cmp); for(int i=0;i<n;i++) { if(a[i].x>k||k<0) { f=0; } else { k=k-a[i].x+a[i].y; } } if(f) puts("yes"); else puts("no"); } return 0; } /* n v //n个礼物 体积为v的背包 ai bi //每个礼物的体积ai 背包对这件礼物的喜爱程度bi(物体放到背包会扩大的体积) 能不能所有礼物都放到背包 【初始体积k=2】 1 2 k= 2-1+2=3 2 3 k= 3-2+3=4 2 1 k= 4-2+1=3 3 1 k= 3-3+1=1 a升序 b降序 */
2017: 开心的cc
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 147 Solved: 26
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Description
Input
Output
Sample Input
2
5
1 0 1 1 0
5
1 1 1 1 1
Sample Output
1
5
HINT
【分析】: 直接看 1比0多的个数 。
思维
Submit: 68 Solved: 23
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![](https://images2017.cnblogs.com/blog/1184092/201712/1184092-20171211145415337-603375901.png)
【代码】:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; int t; int n,x,cnt; int main() { scanf("%d",&t); while(t--) { cnt=0; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&x); if(x==1) cnt++; else cnt--; } printf("%d\n",max(cnt,0)); } return 0; }
不忘初心,砥砺前行!
2021: 剪纸
Time Limit: 4 Sec Memory Limit: 128 MBSubmit: 68 Solved: 23
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Description
Input
Output
Sample Input
1
4
Sample Output
11
【分析】:蓝桥杯原题改编了一点。
DFS
Submit: 519 Solved: 59
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第八届 蓝桥杯 方格分割
【代码】:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <algorithm> #include <string.h> #include <iostream> #include <stdio.h> #include <string> #include <vector> #include <queue> #include <map> #include <set> using namespace std; using namespace std; int ans = 0; int mpt[20+1][20+1]; int N; int dir[4][2] = {0,1,1,0,0,-1,-1,0}; void dfs(int x,int y) { if(x == 0 || y == 0 || x == N || y == N){ ans ++; return; } for(int i = 0 ; i < 4 ; i ++) { int tx = x + dir[i][0]; int ty = y + dir[i][1]; if(mpt[tx][ty])continue; mpt[tx][ty] = 1; mpt[N-tx][N-ty] = 1; dfs(tx,ty); mpt[tx][ty] = 0; mpt[N-tx][N-ty] = 0; } } int main() { int pp;scanf("%d",&pp); while (pp--) { scanf("%d",&N); ans=0;//注意多组数据置位 memset(mpt,0,sizeof(mpt)); mpt[N/2][N/2] = 1; dfs(N/2,N/2); printf("%d\n",ans/4); } return 0; }
2014: 一生之敌
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 519 Solved: 59
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Description
Input
第一行输入一个整数T,表示数据组数。
每组数据输入一个整数n。
1 <= T <= 100000
0 <= n <= 10^19
保证结果存在
Output
输出一个整数。
Sample Input
3
2
6
100
Sample Output
6
6
114
【分析】:这道题实际上就是找2a为为完全平方数的时候, 然后把这些数存起来 (预处理),二分就行了 ,注意用ULL
【代码】:
预处理+二分
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; const int maxn = 1400000 + 10; ULL f[maxn]; void init() { for(int i=0;i<maxn;i++) { f[i] = 4ull * i * i * i + 2ull * i; } } int main() { init(); int t; scanf("%d",&t); while(t--) { ULL n;scanf("%llu",&n); LL ans = lower_bound(f,f+maxn,n) - f; printf("%llu\n",f[ans]); } }