回答編集履歴
2
追記
test
CHANGED
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@@ -193,3 +193,67 @@
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競プロには不慣れなので、無駄な処理をいっぱい書いていると思いますが。
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---
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自由に解いて良いなら、こう。
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```C++
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#include <cstdio>
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int main(void) {
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int N, Y;
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if(std::scanf("%d %d", &N, &Y) != 2) return 1;
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int D = Y / 1000 - N;
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for(int a = 0; a <= N; ++a) {
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if(10000 * a > Y) break;
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int tmp = D - 9*a;
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if(tmp % 4 == 0) {
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int b = tmp / 4;
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int c = N - a - b;
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if(b < 0) continue;
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if(c < 0) continue;
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std::printf("%d %d %d\n", a, b, c);
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return 0;
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}
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}
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std::printf("-1 -1 -1\n");
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}
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```
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[提出結果](https://atcoder.jp/contests/abc085/submissions/4349980)
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1
追記
test
CHANGED
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@@ -23,3 +23,173 @@
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また、ここのans_xyzとmain関数のans_xyzは**全くの別物**です。
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変数の使い方及びポインタの使い方を再確認しましょう。
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書いてみた
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---
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TLEを回避するのにそれなりに苦労しましたが、再帰でAC取りました。
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```C++
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#include <iostream>
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struct Answer {
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int yen_10000;
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int yen_5000;
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int yen_1000;
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bool is_valid() {
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return yen_10000 != -1;
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}
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void print() {
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std::printf(
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"%d %d %d\n",
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yen_10000, yen_5000, yen_1000
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);
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}
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};
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struct Cond {
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bool use_10000 = true;
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bool use_5000 = true;
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bool use_1000 = true;
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};
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// n枚で1000*y円作る組み合わせを返す
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Answer solve(const int n, const int ky, Cond cond) {
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//
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if(n == 0 && ky == 0) {
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return {0, 0, 0};
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}
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if(ky == 0) {
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return {-1, -1, -1};
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}
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if(ky == n) {
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return {0, 0, n};
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}
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//
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if(cond.use_10000) {
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for(int num = ky / 10; num >= 0; --num) {
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auto ans = solve(
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n - num, ky - num*10,
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{false, true, true}
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);
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if(ans.is_valid()) {
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ans.yen_10000 += num;
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return ans;
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}
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}
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}
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if(cond.use_5000) {
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for(int num = ky / 5; num >= 0; --num) {
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auto ans = solve(
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n - num, ky - num*5,
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{false, false, true}
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);
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if(ans.is_valid()) {
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ans.yen_5000 += num;
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return ans;
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160
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161
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}
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162
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}
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}
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167
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168
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return {-1, -1, -1};
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170
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171
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}
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172
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173
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174
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175
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int main(void) {
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177
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int N, Y;
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178
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std::cin >> N >> Y;
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180
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181
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182
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183
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solve(N, Y/1000, {}).print();
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184
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185
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}
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186
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187
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```
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188
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189
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190
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191
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[提出結果](https://atcoder.jp/contests/abc085/submissions/4349569)
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192
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193
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194
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195
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+
競プロには不慣れなので、無駄な処理をいっぱい書いていると思いますが。
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