Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

A  — Area

Time Limit: 1 sec
Memory Limit: 32 MB

Steve has a dream to move from dormitory where we live now to a private house. He has already earned some money by solving various problems and now wants to make a first move towards his dream. Steve is going to buy a plot for his future house. However, due to high tax rate, he is going to buy only one plot. Certainly Steve wants to maximize its area for the amount of money he owns (and if there are several plots of same area, Steve certainly chooses a cheaper one). Can you help Steve?

The whole district where Steve wants to reside is divided into N * M equal strips. Each strip is sold separately, for its own price P_i;j. According to the country law, plot is defined as a rectangular collection of strips with sides parallel to the district.

INPUT

There is a number of tests T (T ≤ 110) on the first line. After T tests follows. Each test case is defined by three numbers N M K (N; M ≤ 100; K ≤ 10⁹). Next N lines with M numbers each stands for prices of the strips P_i;j (P_i;j ≤ 10⁶).

OUTPUT

For each test print a line formatted as "Case #T: S P", where T stands for test case number (starting from 1), S for maximal plot area that Steve can afford and P for plot’s total cost.

SAMPLE INPUT

1
5 6 15
10 1 10 20 10 10
30 1 1 5 1 1
50 1 1 20 1 1
10 5 5 10 5 1
40 10 90 1 10 10

SAMPLE OUTPUT

Case #1: 6 10

---

Problem by: Aleksej Viktorchik; Leonid Sislo
Huge Easy Contest #2

O(n^3)


#include <stdio.h>
inline int readchar() {
    const int N = 1048576;
    static char buf[N];
    static char *p = buf, *end = buf;
    if(p == end) {
        if((end = buf + fread(buf, 1, N, stdin)) == buf) return EOF;
        p = buf;
    }
    return *p++;
}
inline int ReadInt(int *x) {
    static char c, neg;
    while((c = readchar()) < '-')    {if(c == EOF) return 0;}
    neg = (c == '-') ? -1 : 1;
    *x = (neg == 1) ? c-'0' : 0;
    while((c = readchar()) >= '0')
        *x = (*x << 3) + (*x << 1) + c-'0';
    *x *= neg;
    return 1;
}
int main() {
    int t, N, M, K;
    int g[105][105];
    int cases = 0;
    int i, j, k, l;
    //scanf("%d", &t);
    ReadInt(&t);
    while(t--) {
        //scanf("%d %d %d", &N, &M, &K);
        ReadInt(&N);
        ReadInt(&M);
        ReadInt(&K);
        for(i = 0; i < N; i++)
            for(j = 0; j < M; j++)
                //scanf("%d", &g[i][j]);
                ReadInt(&g[i][j]);
        int mx = 0, mxv = 0;
        long long tmp;
        int cnt = 0;
        for(i = 0; i < N; i++) {
            int sum[105] = {};
            for(j = i; j < N; j++) {
                for(k = 0, l = 0; k < M; k++) {
                    sum[k] += g[j][k];
                    if(k == 0)  tmp = 0;
                    tmp += sum[k];
                    while(tmp > K)
                        tmp -= sum[l++];
                    if((j-i+1)*(k-l+1) > mx)
                        mx = (j-i+1)*(k-l+1), mxv = K;
                    if((j-i+1)*(k-l+1) == mx && tmp < mxv)
                        mxv = tmp;
                }
            }
        }
        printf("Case #%d: %d %d\n", ++cases, mx, mxv);
    }
    return 0;
}