這一場PP說很簡單，我就閑來無事敲一敲好了，反正也考完了（霧，不過只剩數(shù)據(jù)結(jié)構(gòu)和c++ A. Okabe and Future Gadget Laboratory time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Okabe needs to renovate the?Future Gadget Laboratory?after he tried doing some crazy experiments! The lab is represented as an?n?by?nsquare grid of integers. A?good?lab is defined as a lab in which every number not equal to?1?can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every?x,?y?such that?1?≤?x,?y?≤?n?and?ax,?y?≠?1, there should exist two indices?s?and?t?so that?ax,?y?=?ax,?s?+?at,?y, where?ai,?j?denotes the integer in?i-th row and?j-th column.

Help Okabe determine whether a given lab is?good!

Input

The first line of input contains the integer?n?(1?≤?n?≤?50)?— the size of the lab.

The next?n?lines contain?n?space-separated integers denoting a row of the grid. The?j-th integer in the?i-th row is?ai,?j?(1?≤?ai,?j?≤?105).

Output

Print "Yes" if the given lab is?good?and "No" otherwise.

You can output each letter in upper or lower case.

Examples input 3
1 1 2
2 3 1
6 4 1 output Yes input 3
1 5 2
1 1 1
1 2 3 output No Note

In the first sample test, the?6?in the bottom left corner is valid because it is the sum of the?2?above it and the?4?on the right. The same holds for every number not equal to?1?in this table, so the answer is "Yes".

In the second sample test, the?5?cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".

?這個題的意思就是給你一個矩陣，假如不是1的話讓你找一個所在行的數(shù)和所在列的數(shù)是不是和他想等，正確yes，否則no。暴力就可以的

#include <stdio.h> int a[55][55]; int n; int solve(int x,int y) {for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)if(a[x][i]+a[j][y]==a[x][y]) return 1;return 0; } int main() {scanf("%d",&n);for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)scanf("%d",a[i]+j);for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)if(a[i][j]!=1)if(solve(i,j)==0)return 0*printf("No");return 0*printf("Yes"); }

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B. Okabe and Banana Trees time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.

Consider the point?(x,?y)?in the 2D plane such that?x?and?y?are integers and?0?≤?x,?y. There is a tree in such a point, and it has?x?+?ybananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation?. Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.

Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.

Okabe is sure that the answer does not exceed?1018. You can trust him.

Input

The first line of input contains two space-separated integers?m?and?b?(1?≤?m?≤?1000,?1?≤?b?≤?10000).

Output

Print the maximum number of bananas Okabe can get from the trees he cuts.

Examples input 1 5 output 30 input 2 3 output 25 Note

The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has?30?bananas.

?給你一條經(jīng)過1，3，4象限的直線，讓我找到一個矩形價(jià)值最大的值，一個矩形的價(jià)值是這個矩形中所有的點(diǎn)X和Y的加和，這是個等差數(shù)列的和

直接暴力枚舉好了

#include <stdio.h> #include <bits/stdc++.h> using namespace std; typedef long long LL; int main() {int m,b;scanf("%d%d",&m,&b);LL ma=0;for(int y=0; y<=b; y++) {LL x=(b-y)*m;LL t=x*(x+1)/2*(y+1)+y*(y+1)/2*(x+1);ma=max(t,ma);}printf("%lld\n",ma);return 0; }

?

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轉(zhuǎn)載于:https://www.cnblogs.com/BobHuang/p/7084254.html

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