C - March
Time limit : 2sec / Memory limit : 256MB
Score : 300 points
Problem Statement
There are N people. The name of the i-th person is Si.
We would like to choose three people so that the following conditions are met:
- The name of every chosen person begins with
M,A,R,CorH. - There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
- 1≤N≤105
- Si consists of uppercase English letters.
- 1≤|Si|≤10
- Si≠Sj(i≠j)
Input
Input is given from Standard Input in the following format:
N S1 : SN
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Sample Input 1
5 MASHIKE RUMOI OBIRA HABORO HOROKANAI
Sample Output 1
2
We can choose three people with the following names:
-
MASHIKE,RUMOI,HABORO -
MASHIKE,RUMOI,HOROKANAI
Thus, we have two ways.
Sample Input 2
4 ZZ ZZZ Z ZZZZZZZZZZ
Sample Output 2
0
Note that there may be no ways to choose three people so that the given conditions are met.
Sample Input 3
5 CHOKUDAI RNG MAKOTO AOKI RINGO
Sample Output 3
7
std:
#include <cstdio > #include <iostream > using namespace std; typedef long long ll; string s; int N; ll m,a,r,c,h; ll D[5]; int P[10]={0,0,0,0,0,0,1,1,1,2}; int Q[10]={1,1,1,2,2,3,2,2,3,3}; int R[10]={2,3,4,3,4,4,3,4,4,4}; int main() { scanf("%d",&N); for(int i=0;i<N;i++) { cin>>s; if(s[0]==’M’)m++; if(s[0]==’A’)a++; if(s[0]==’R’)r++; if(s[0]==’C’)c++; if(s[0]==’H’)h++; } D[0]=m,D[1]=a,D[2]=r,D[3]=c,D[4]=h; ll res=0; for(int d=0;d<10;d++) res+=D[P[d]]*D[Q[d]]*D[R[d]]; printf("%lld\n",res); }
暴力枚举每一种情况：共10种
#include<cstdio> #include<cstring> #include<string> #include<iostream> using namespace std; long long n,a[6],ans; string s[200000]; int main() { scanf("%lld",&n); for(int i=1;i<=n;i++) cin>>s[i]; for(int i=1;i<=n;i++){ if(s[i][0]==‘M‘) ++a[1]; if(s[i][0]==‘A‘) ++a[2]; if(s[i][0]==‘R‘) ++a[3]; if(s[i][0]==‘C‘) ++a[4]; if(s[i][0]==‘H‘) ++a[5]; } ans+=a[1]*a[2]*a[3]; ans+=a[1]*a[2]*a[4]; ans+=a[1]*a[2]*a[5]; ans+=a[1]*a[3]*a[4]; ans+=a[1]*a[3]*a[5]; ans+=a[1]*a[4]*a[5]; ans+=a[2]*a[3]*a[4]; ans+=a[2]*a[3]*a[5]; ans+=a[3]*a[4]*a[5]; ans+=a[2]*a[4]*a[5]; printf("%lld\n",ans); return 0; }
D - Practical Skill Test
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
Problem Statement
We have a grid with H rows and W columns. The square at the i-th row and the j-th column will be called Square (i,j).
The integers from 1 through H×W are written throughout the grid, and the integer written in Square (i,j) is Ai,j.
You, a magical girl, can teleport a piece placed on Square (i,j) to Square (x,y) by consuming |x?i|+|y?j| magic points.
You now have to take Q practical tests of your ability as a magical girl.
The i-th test will be conducted as follows:
-
Initially, a piece is placed on the square where the integer Li is written.
-
Let x be the integer written in the square occupied by the piece. Repeatedly move the piece to the square where the integer x+D is written, as long as x is not Ri. The test ends when x=Ri.
-
Here, it is guaranteed that Ri?Li is a multiple of D.
For each test, find the sum of magic points consumed during that test.
Constraints
- 1≤H,W≤300
- 1≤D≤H×W
- 1≤Ai,j≤H×W
- Ai,j≠Ax,y((i,j)≠(x,y))
- 1≤Q≤105
- 1≤Li≤Ri≤H×W
- (Ri?Li) is a multiple of D.
Input
Input is given from Standard Input in the following format:
H W D A1,1 A1,2 … A1,W : AH,1 AH,2 … AH,W Q L1 R1 : LQ RQ
Output
For each test, print the sum of magic points consumed during that test.
Output should be in the order the tests are conducted.
Sample Input 1
3 3 2 1 4 3 2 5 7 8 9 6 1 4 8
Sample Output 1
5
-
4 is written in Square (1,2).
-
6 is written in Square (3,3).
-
8 is written in Square (3,1).
Thus, the sum of magic points consumed during the first test is (|3?1|+|3?2|)+(|3?3|+|1?3|)=5.
Sample Input 2
4 2 3 3 7 1 4 5 2 6 8 2 2 2 2 2
Sample Output 2
0 0
Note that there may be a test where the piece is not moved at all, and there may be multiple identical tests.
Sample Input 3
5 5 4 13 25 7 15 17 16 22 20 2 9 14 11 12 1 19 10 6 23 8 18 3 21 5 24 4 3 13 13 2 10 13 13
Sample Output 3
0 5 0
std:DP预处理
#include <cstdio > #define abs(x) ((x>0)?x:(-(x))) int H,W,D,A; int Q,L,R; int px[90001],py[90001]; int d[90001]; int main() { scanf("%d%d%d",&H,&W,&D); for(int i=0;i<H;i++) { for(int j=0;j<W;j++) { scanf("%d",&A); px[A]=i,py[A]=j; } } for(int i=D+1;i<=H*W;i++) d[i]=d[i-D]+abs(px[i]-px[i-D])+abs(py[i]-py[i-D]); scanf("%d",&Q); while(Q--) { scanf("%d%d",&L,&R); printf("%d\n",d[R]-d[L]); } }
说明一下：A B C D 共四题
由于A B 太水，且没有可取之处，所以不在博客显示
网址：https://abc089.contest.atcoder.jp/
